Solve Using the Quadratic Formula x^2+18x+81=25

$x^{2} + 18 x + 81 = 25$

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $25$ from both sides of the equation.

$x^{2} + 18 x + 81 - 25 = 0$

Subtract $25$ from $81$ .

$x^{2} + 18 x + 56 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 18$ , and $c = 56$ into the quadratic formula and solve for $x$ .

$\frac{- 18 \pm \sqrt{18^{2} - 4 \cdot (1 \cdot 56)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $18$ to the power of $2$ .

$x = \frac{- 18 \pm \sqrt{324 - 4 \cdot 1 \cdot 56}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 56$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 18 \pm \sqrt{324 - 4 \cdot 56}}{2 \cdot 1}$

Multiply $- 4$ by $56$ .

$x = \frac{- 18 \pm \sqrt{324 - 224}}{2 \cdot 1}$

Subtract $224$ from $324$ .

$x = \frac{- 18 \pm \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 18 \pm \sqrt{10^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 18 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 18 \pm 10}{2}$

Simplify $\frac{- 18 \pm 10}{2}$ .

$x = - 9 \pm 5$

Step 5

The final answer is the combination of both solutions.

$x = - 4, - 14$