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Example

Solving x26x+5=0x^2 - 6x + 5 = 0 Using the Quadratic Formula

Solve x26x+5=0x^2 - 6x + 5 = 0 by applying the Quadratic Formula. This example demonstrates the procedure when the leading coefficient is 11 and the linear coefficient is negative.

Step 1 — Identify aa, bb, cc. The equation is already in standard form. Here a=1a = 1, b=6b = -6, and c=5c = 5.

Step 2 — Substitute into the Quadratic Formula:

x=(6)±(6)24(1)(5)2(1)x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)}

Step 3 — Simplify. The double negative gives (6)=6-(-6) = 6. Inside the square root: (6)2=36(-6)^2 = 36 and 4(1)(5)=204(1)(5) = 20, so 3620=1636 - 20 = 16. Since 16=4\sqrt{16} = 4:

x=6±42x = \frac{6 \pm 4}{2}

Split into two solutions:

x=6+42=102=5orx=642=22=1x = \frac{6 + 4}{2} = \frac{10}{2} = 5 \qquad \text{or} \qquad x = \frac{6 - 4}{2} = \frac{2}{2} = 1

Step 4 — Check both solutions:

For x=1x = 1: (1)26(1)+5=16+5=0(1)^2 - 6(1) + 5 = 1 - 6 + 5 = 0

For x=5x = 5: (5)26(5)+5=2530+5=0(5)^2 - 6(5) + 5 = 25 - 30 + 5 = 0

The solutions are x=5x = 5 and x=1x = 1. When bb is negative, the first term in the numerator of the Quadratic Formula becomes positive because negating a negative number yields a positive result — here, (6)=6-(-6) = 6. Careful handling of this double negative is essential to avoid sign errors.

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Updated 2026-04-21

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