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How to Solve Quadratic Equations: 3 Methods with Examples

Published 14 March 2026 · 10 min read

A quadratic equation is any equation that can be written in the standard form ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants and a≠0a \neq 0. These equations appear throughout mathematics, physics, engineering, and economics. Whether you are modelling projectile motion, optimising profit, or solving geometry problems, quadratic equations are essential.

There are three main methods for solving quadratics: the quadratic formula, factoring, and completing the square. Each method has its strengths. In this guide, you will learn all three approaches with fully worked examples so you can choose the right tool for any problem.

Key Concepts

Standard Form

Every quadratic equation can be rearranged into standard form:

ax2+bx+c=0ax^2 + bx + c = 0

The coefficient aa is the quadratic coefficient, bb is the linear coefficient, and cc is the constant term.

The Discriminant

The discriminant Δ=b2−4ac\Delta = b^2 - 4ac tells you how many real solutions exist:

  • Δ>0\Delta > 0: two distinct real solutions
  • Δ=0\Delta = 0: one repeated real solution
  • Δ<0\Delta < 0: no real solutions (two complex solutions)

Method 1: The Quadratic Formula

The quadratic formula works for every quadratic equation. It is derived by completing the square on the general form and gives:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Example 1: Solve 2x2+5x−3=02x^2 + 5x - 3 = 0

Identify: a=2a = 2, b=5b = 5, c=−3c = -3.

x=−5±52−4(2)(−3)2(2)x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}
x=−5±25+244x = \frac{-5 \pm \sqrt{25 + 24}}{4}
x=−5±494x = \frac{-5 \pm \sqrt{49}}{4}
x=−5±74x = \frac{-5 \pm 7}{4}

So x=24=12x = \frac{2}{4} = \frac{1}{2} or x=−124=−3x = \frac{-12}{4} = -3.

Example 2: Solve x2−4x+4=0x^2 - 4x + 4 = 0

Here a=1a = 1, b=−4b = -4, c=4c = 4.

x=4±16−162=4±02x = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2}

The discriminant is zero, so there is one repeated solution: x=2x = 2.

Method 2: Factoring

Factoring is the quickest method when it works. The idea is to rewrite the quadratic as a product of two binomials. If ax2+bx+c=0ax^2 + bx + c = 0 factors as (px+q)(rx+s)=0(px + q)(rx + s) = 0, then by the zero product property, either px+q=0px + q = 0 or rx+s=0rx + s = 0.

Example 3: Solve x2+5x+6=0x^2 + 5x + 6 = 0

Find two numbers that multiply to 6 and add to 5. Those numbers are 2 and 3.

x2+5x+6=(x+2)(x+3)=0x^2 + 5x + 6 = (x + 2)(x + 3) = 0

So x=−2x = -2 or x=−3x = -3.

Example 4: Solve 3x2−12x=03x^2 - 12x = 0

Factor out the greatest common factor first:

3x(x−4)=03x(x - 4) = 0

So x=0x = 0 or x=4x = 4.

Method 3: Completing the Square

Completing the square transforms the equation into the form (x+h)2=k(x + h)^2 = k, which you can solve by taking square roots. This method always works and is particularly useful for deriving the vertex form of a parabola.

Step-by-step process

  1. Move the constant term to the right side of the equation.
  2. If a≠1a \neq 1, divide every term by aa.
  3. Take half the coefficient of xx, square it, and add it to both sides.
  4. Factor the left side as a perfect square trinomial.
  5. Take the square root of both sides and solve for xx.

Example 5: Solve x2+6x+2=0x^2 + 6x + 2 = 0

Move the constant:

x2+6x=−2x^2 + 6x = -2

Half of 6 is 3, and 32=93^2 = 9. Add 9 to both sides:

x2+6x+9=−2+9x^2 + 6x + 9 = -2 + 9
(x+3)2=7(x + 3)^2 = 7

Take the square root:

x+3=±7x + 3 = \pm\sqrt{7}
x=−3±7x = -3 \pm \sqrt{7}

So x≈−0.354x \approx -0.354 or x≈−5.646x \approx -5.646.

Example 6: Solve 2x2−8x+1=02x^2 - 8x + 1 = 0

Divide by 2: x2−4x+12=0x^2 - 4x + \frac{1}{2} = 0

Move the constant:

x2−4x=−12x^2 - 4x = -\frac{1}{2}

Half of −4-4 is −2-2, and (−2)2=4(-2)^2 = 4. Add 4 to both sides:

(x−2)2=72(x - 2)^2 = \frac{7}{2}
x=2±72x = 2 \pm \sqrt{\frac{7}{2}}

So x≈3.871x \approx 3.871 or x≈0.129x \approx 0.129.

Choosing the Right Method

MethodBest whenLimitation
Quadratic formulaAlways works, messy coefficientsMore computation
FactoringSmall integer coefficientsNot always factorable over integers
Completing the squareVertex form needed, deriving formulasMore steps than factoring

Common Mistakes

Forgetting the negative root

The ±\pm in the quadratic formula means there are usually two solutions. Always compute both.

Sign errors in the discriminant

Be careful with b2−4acb^2 - 4ac when bb or cc is negative. A negative times a negative is positive.

Not rearranging to standard form

Before applying any method, make sure the equation equals zero. Move all terms to one side first.

Dividing by xx and losing a solution

If you have 3x2=12x3x^2 = 12x, do not divide both sides by xx. You will lose the solution x=0x = 0. Factor instead.

Try It Yourself

Ready to practise? Use our free quadratic formula calculator to check your work instantly. Enter your coefficients and see the full step-by-step solution.

Open Quadratic Formula Calculator

For general equation solving, try our Solve For X Calculator.

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