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Solving Quadratic Equations – Step by Step

A guided lesson on solving quadratic equations using factoring, completing the square, and the quadratic formula.

Last updated: 9 March 2026

β˜…β˜…β˜†β˜†β˜† Level 2AlgebraLessonQuadraticFactoringQuadratic Formula4 min read

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What is a Quadratic Equation?

A quadratic equation has degree 2:

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

There are three main methods to solve them.

Method 1: Factoring

If the quadratic factors nicely:

x2βˆ’5x+6=(xβˆ’2)(xβˆ’3)=0x^2 - 5x + 6 = (x-2)(x-3) = 0

So x=2x = 2 or x=3x = 3.

  1. 1

    Write in standard form

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

  2. 2

    Find two numbers that multiply to c and add to b

    For x2βˆ’5x+6x^2 - 5x + 6: need pβ‹…q=6p \cdot q = 6 and p+q=βˆ’5p + q = -5. So p=βˆ’2,q=βˆ’3p = -2, q = -3.

  3. 3

    Factor and solve

    (xβˆ’2)(xβˆ’3)=0β€…β€ŠβŸΉβ€…β€Šx=2(x - 2)(x - 3) = 0 \implies x = 2 or x=3x = 3.

Method 2: The Quadratic Formula

Always works:

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Solve xΒ² + 2x βˆ’ 8 = 0 using the formula

a=1,b=2,c=βˆ’8a=1, b=2, c=-8

Ξ”=4+32=36\Delta = 4 + 32 = 36

x=βˆ’2Β±62x = \frac{-2 \pm 6}{2}: x1=2x_1 = 2, x2=βˆ’4x_2 = -4

Method 3: Completing the Square

Useful for understanding the vertex form a(xβˆ’h)2+ka(x-h)^2 + k.

  1. 1

    Move constant to the right

    x2+4x=5x^2 + 4x = 5

  2. 2

    Add (b/2)Β² to both sides

    x2+4x+4=5+4=9x^2 + 4x + 4 = 5 + 4 = 9

  3. 3

    Factor and solve

    (x+2)2=9β€…β€ŠβŸΉβ€…β€Šx+2=Β±3β€…β€ŠβŸΉβ€…β€Šx=1(x+2)^2 = 9 \implies x + 2 = \pm 3 \implies x = 1 or x=βˆ’5x = -5

Tip

If the discriminant b2βˆ’4ac<0b^2 - 4ac < 0, the equation has no real solutions.

Real world: ball thrown into the air

A ball is launched upward with the height (in metres) modelled by h=βˆ’5t2+20th = -5t^2 + 20t, where tt is time in seconds. Setting h=0h = 0 to find when it lands: βˆ’5t(tβˆ’4)=0-5t(t - 4) = 0, so t=0t = 0 or t=4t = 4 seconds. Quadratic equations describe every projectile β€” from basketballs to rockets.

Real world: satellite dish shape

A parabolic satellite dish has a cross-section described by y=0.1x2y = 0.1x^2. To find where the dish is 5 m wide at height y=0.9y = 0.9 m: 0.9=0.1x2β‡’x2=9β‡’x=Β±30.9 = 0.1x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3 m. Engineers use quadratic equations to design the exact curvature that focuses signals onto the receiver.

Real world: fencing a rectangular garden

A gardener has 40 m of fence and wants to enclose a rectangular area of 96 mΒ². Setting up: 2l+2w=402l + 2w = 40 so w=20βˆ’lw = 20 - l, then l(20βˆ’l)=96β‡’l2βˆ’20l+96=0β‡’(lβˆ’8)(lβˆ’12)=0l(20-l) = 96 \Rightarrow l^2 - 20l + 96 = 0 \Rightarrow (l-8)(l-12) = 0. The garden is 8Β mΓ—12Β m8\text{ m} \times 12\text{ m}.

Tip Before choosing a method, check the discriminant b2βˆ’4acb^2 - 4ac. If it is a perfect square, factoring will be quick. Otherwise use the quadratic formula.

Practice Problems

Factor: xΒ² βˆ’ 7x + 12 = 0

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(xβˆ’3)(xβˆ’4) = 0 β†’ x = 3 or x = 4

Use quadratic formula: xΒ² βˆ’ 4x + 3 = 0

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Ξ” = 16βˆ’12 = 4. x = (4Β±2)/2 β†’ x = 3 or x = 1

Complete the square: xΒ² + 6x βˆ’ 7 = 0

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(x+3)Β² = 16 β†’ x+3 = Β±4 β†’ x = 1 or x = βˆ’7