How to Solve Quadratic Equations — 4 Fastest Methods (With Examples)

What is a quadratic equation?

A quadratic equation (or second-degree equation) has the general form:

$a{x}^2+bx+c=0$

Where a, b, c are real coefficients, and the required condition is a ≠ 0. If a = 0, the equation becomes linear or trivially true/false depending on the values of b and c.

Quadratic equations appear at almost every level of math education — from grade 9 basic curriculum to applications in Physics, Economics, and Engineering. In the national high school exam, solving quadratic equations often appears as an intermediate step in more complex problems: finding roots of trigonometric equations, finding conditions for a graph to cross the x-axis, or solving real-world problems.

What makes quadratic equations interesting is that there are 4 different methods to solve them — each method is optimal for a specific type of problem. Mastering all 4 methods and knowing when to use each one will help you solve faster and make significantly fewer mistakes.

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Method 1 — Quadratic Formula (Delta)

This is the most general method — applicable to every quadratic equation, regardless of the coefficients.

The Discriminant Delta

From the equation ax² + bx + c = 0 (a ≠ 0), compute the discriminant:

$\Delta =b^2-4ac$

Three Cases of Delta

Condition	Number of Roots	Formula
Delta > 0	Two distinct roots	x = (−b ± √Delta) / (2a)
Delta = 0	Double root	x₁ = x₂ = −b / (2a)
Delta < 0	No real roots	No real solution exists

Written as a mathematical formula:

$x_{1,2}=\frac{-b\pm \sqrt{\Delta }}{2a}(\text{when }\Delta >0)$

Reduced Discriminant Formula (Delta')

When b is even (let b = 2b'), we can use the reduced discriminant:

${\Delta }^{\prime }=b^{\prime 2}-ac$

Reduced quadratic formula:

$x_{1,2}=\frac{-b^{\prime }\pm \sqrt{{\Delta }^{\prime }}}{a}$

Example: the equation 2x² − 6x + 4 = 0 has b = −6 (even), so b' = −3, Delta' = 9 − 8 = 1, roots are x₁ = 2 and x₂ = 1.

Applying the Delta Formula

Example 1: Solve x² − 5x + 6 = 0.

Step 1: Identify a = 1, b = −5, c = 6.

Step 2: Calculate Delta = (−5)² − 4·1·6 = 25 − 24 = 1. Delta > 0 so two distinct roots.

Step 3: Find roots:

$x_{1,2}=\frac{5\pm \sqrt{1}}{2}=\frac{5\pm 1}{2}$

So x₁ = 3 and x₂ = 2.

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Method 2 — Vieta's Formulas

Vieta's formulas (or Vieta's theorem) describe the relationship between the roots and coefficients of a quadratic equation.

Statement of Vieta's Formulas

If x₁, x₂ are the two roots of ax² + bx + c = 0, then:

$x_1+x_2=-\frac{b}{a}$

$x_1\cdot x_2=\frac{c}{a}$

Application: Guessing Roots When Sum and Product Are Known

Vieta's formulas allow us to "guess" roots when we know their sum and product. Example: the equation x² − 7x + 12 = 0 has sum of roots S = 7 and product of roots P = 12. We need two numbers with sum 7 and product 12 — those are 3 and 4. So x₁ = 3, x₂ = 4 without calculating Delta!

Converse of Vieta's Formulas

If two numbers x₁, x₂ satisfy x₁ + x₂ = S and x₁·x₂ = P, then they are the roots of:

$x^2-Sx+P=0$

Example 2: Find the quadratic equation with roots x₁ = −2 and x₂ = 5.

S = −2 + 5 = 3, P = (−2)·5 = −10.

The required equation: x² − 3x − 10 = 0.

Vieta Problems in Exams

A very common type of problem: "Given the equation x² − mx + m − 1 = 0 has two positive roots. Find m."

Conditions for two positive roots:

• Delta ≥ 0: (m − 2)² ≥ 0 — always true.
• x₁ + x₂ = m > 0.
• x₁·x₂ = m − 1 > 0, so m > 1.

Therefore the condition is m > 1.

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Method 3 — Special Root Inspection

This is an extremely fast method for certain special forms of equations — saving significant time in exams.

Case a + b + c = 0

If the sum of all coefficients equals zero, then x = 1 is always a root.

The other root: x₂ = c/a (from Vieta's: x₁·x₂ = c/a, x₁ = 1).

Example 3: Solve 3x² − 7x + 4 = 0.

Check: a + b + c = 3 − 7 + 4 = 0 ✓

So x₁ = 1 and x₂ = c/a = 4/3.

Case a − b + c = 0

If a − b + c = 0 (equivalent to f(−1) = 0), then x = −1 is always a root.

The other root: x₂ = −c/a.

Example 4: Solve 5x² + 3x − 8 = 0.

Check: a + b + c = 5 + 3 − 8 = 0 ✓. So x₁ = 1 and x₂ = c/a = −8/5.

Case c = 0

If c = 0, the equation becomes ax² + bx = 0, factor out x:

$x(ax+b)=0\Rightarrow x=0\text{ or }x=-\frac{b}{a}$

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Method 4 — Substitution

The substitution method is used to reduce higher-degree equations to simple quadratic equations.

Biquadratic Equations

A biquadratic equation has the form: ax⁴ + bx² + c = 0 (no odd powers of x).

Method: Let t = x² (t ≥ 0), the equation becomes:

$a{t}^2+bt+c=0$

Solve the quadratic in t, then find x from t.

Example 5: Solve x⁴ − 5x² + 4 = 0.

Let t = x² (t ≥ 0):

$t^2-5t+4=0\Rightarrow (t-1)(t-4)=0$

• t = 1 → x² = 1 → x = ±1
• t = 4 → x² = 4 → x = ±2

So the equation has 4 roots: x ∈ {−2, −1, 1, 2}.

Other Substitutions

This technique also applies to trigonometric, exponential, and logarithmic equations:

• Exponential: 4^x − 5·2^x + 4 = 0 → let t = 2^x → t² − 5t + 4 = 0.
• Trigonometric: sin²(x) − 3·sin(x) + 2 = 0 → let t = sin(x), condition −1 ≤ t ≤ 1.

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5 Step-by-Step Practice Problems

Problem 1 — Using Delta

Solve 2x² + 7x − 15 = 0.

Solution:

a = 2, b = 7, c = −15.

$\Delta =49+120=169=13^2$

$x_{1,2}=\frac{-7\pm 13}{4}$

x₁ = 6/4 = 3/2, x₂ = −20/4 = −5.

Problem 2 — Using Vieta's to Guess Roots

Solve x² − 11x + 28 = 0.

Solution:

Sum S = 11, product P = 28. Find two numbers with sum 11 and product 28: those are 4 and 7.

x₁ = 4, x₂ = 7.

Problem 3 — Special Root Inspection

Solve 7x² − 13x + 6 = 0.

Solution:

a + b + c = 7 − 13 + 6 = 0 ✓

x₁ = 1, x₂ = c/a = 6/7.

Problem 4 — Advanced Vieta Problem

Given the equation x² − 3x + k = 0 with two roots x₁, x₂. Calculate x₁² + x₂².

Solution:

By Vieta's: x₁ + x₂ = 3 and x₁·x₂ = k.

$x_1^2+x_2^2=(x_1+x_2{)}^2-2x_1{x}_2=9-2k$

Problem 5 — Biquadratic Equation

Solve 9x⁴ − 10x² + 1 = 0.

Solution:

Let t = x² (t ≥ 0): 9t² − 10t + 1 = 0.

$\Delta =100-36=64,\sqrt{\Delta }=8$

t₁ = (10 + 8)/18 = 1, t₂ = (10 − 8)/18 = 1/9.

• t = 1 → x = ±1
• t = 1/9 → x = ±1/3

So x ∈ {−1, −1/3, 1/3, 1}.

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Quick Guide to Choosing the Fastest Method

When facing a quadratic equation, check in this order:

1. c = 0? → Factor out x: x(ax + b) = 0 → done in 10 seconds.
2. a + b + c = 0? → Instantly x₁ = 1, x₂ = c/a.
3. a − b + c = 0? → Instantly x₁ = −1, x₂ = −c/a.
4. Integer coefficients, |c/a| small? → Try Vieta's: find two integers with sum = −b/a and product = c/a.
5. None of the above apply → Use the quadratic formula (always works).

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Frequently Asked Questions

Q1: What is a double root and when does it occur?

A double root (or root of multiplicity 2) is when the equation has exactly one real root x₁ = x₂, which occurs when Delta = 0. Example: x² − 4x + 4 = 0 = (x−2)² = 0 has double root x = 2. On the graph, this is where the parabola is tangent to the x-axis (touches but doesn't cross).

Q2: When does a quadratic equation have no solution?

The equation ax² + bx + c = 0 has no real solution when Delta = b² − 4ac < 0. This means the parabola y = ax² + bx + c does not intersect the x-axis. In terms of complex numbers, the equation still has 2 complex conjugate roots, but in the standard high school curriculum we say "the equation has no solution".

Q3: When is the converse of Vieta's formulas used in exams?

The converse of Vieta's formulas typically appears in problems like: "Find m such that the equation has two roots satisfying the condition..." For example: "Find m such that the two roots x₁, x₂ of x² − mx + 2m − 1 = 0 satisfy x₁² + x₂² = 7." This is a reverse Vieta problem — using the root condition to find the parameter m.

Q4: What is Delta' and how does it relate to Delta?

Delta' = b'² − ac is the reduced discriminant used when b is even (b = 2b'). The relationship: Delta = 4·Delta'. Therefore Delta > 0 if and only if Delta' > 0, and similarly for the other cases. The reduced formula x = (−b' ± √Delta') / a is more convenient because it avoids calculating large numbers when b is large.

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Mastering the 4 methods for solving quadratic equations is the first step toward confidence with higher-level problems — from trigonometric equations and exponential-logarithmic equations to optimization problems. If you want to practice solving quadratic equations with immediate feedback, the Witza app provides thousands of problems organized by type with detailed step-by-step solutions — perfect for daily practice.