Solve Using the Quadratic Formula x^2+x=-9x-16

$x^{2} + x = - 9 x - 16$

Step 1

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

Move all the expressions to the left side of the equation.

Add $9 x$ to both sides of the equation.

$x^{2} + x + 9 x = - 16$

Add $16$ to both sides of the equation.

$x^{2} + x + 9 x + 16 = 0$

Add $x$ and $9 x$ .

$x^{2} + 10 x + 16 = 0$

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 $a = 1$ , $b = 10$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 16$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 16}}{2 \cdot 1}$

Multiply $- 4$ by $16$ .

$x = \frac{- 10 \pm \sqrt{100 - 64}}{2 \cdot 1}$

Subtract $64$ from $100$ .

$x = \frac{- 10 \pm \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 10 \pm \sqrt{6^{2}}}{2 \cdot 1}$

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

$x = \frac{- 10 \pm 6}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 10 \pm 6}{2}$

Simplify $\frac{- 10 \pm 6}{2}$ .

$x = - 5 \pm 3$

Step 5

The final answer is the combination of both solutions.

$x = - 2, - 8$