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ALGEBRA · QUADRATICS

The Quadratic Formula

One formula that solves every quadratic — and tells you how many solutions exist.

13 min read 3 practice questions
When a quadratic will not factor nicely, the quadratic formula always works. It comes from completing the square on ax2+bx+c=0ax^2 + bx + c = 0.
x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
The expression under the root, D=b2−4acD = b^2 - 4ac, is the discriminant. If D>0D > 0 there are two real solutions, if D=0D = 0 there is exactly one, and if D<0D < 0 there are no real solutions.

Worked example · Using the formula

Solve 2x2−4x−6=02x^2 - 4x - 6 = 0.
  1. 1
    Identify a=2a = 2, b=−4b = -4, c=−6c = -6.
  2. 2
    Discriminant: (−4)2−4(2)(−6)=16+48=64(-4)^2 - 4(2)(-6) = 16 + 48 = 64.
  3. 3
    x=4±84x = \frac{4 \pm 8}{4}, so x=3x = 3 or x=−1x = -1.

CHECK YOUR UNDERSTANDING

Practice

Answer each question, then check.

1
Find the discriminant of x2+6x+5x^2 + 6x + 5.
2
How many real solutions does x2+2x+5=0x^2 + 2x + 5 = 0 have?
3
What is the larger solution of x2−2x−15=0x^2 - 2x - 15 = 0?

Numeric answers accept decimals or fractions like 3/4.