Solve Using the Quadratic Formula x^2+12x+36=25

$x^{2} + 12 x + 36 = 25$

Step 1

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

Subtract $25$ from both sides of the equation.

$x^{2} + 12 x + 36 - 25 = 0$

Subtract $25$ from $36$ .

$x^{2} + 12 x + 11 = 0$

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 $a = 1$ , $b = 12$ , and $c = 11$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot 11)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot 11}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 11$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 11}}{2 \cdot 1}$

Multiply $- 4$ by $11$ .

$x = \frac{- 12 \pm \sqrt{144 - 44}}{2 \cdot 1}$

Subtract $44$ from $144$ .

$x = \frac{- 12 \pm \sqrt{100}}{2 \cdot 1}$

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

$x = \frac{- 12 \pm \sqrt{10^{2}}}{2 \cdot 1}$

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

$x = \frac{- 12 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 12 \pm 10}{2}$

Simplify $\frac{- 12 \pm 10}{2}$ .

$x = - 6 \pm 5$

Step 5

The final answer is the combination of both solutions.

$x = - 1, - 11$