Solve Using the Quadratic Formula x^2+4x+17=8-2x

$x^{2} + 4 x + 17 = 8 - 2 x$

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.

Subtract $8$ from both sides of the equation.

$x^{2} + 4 x + 17 - 8 = - 2 x$

Add $2 x$ to both sides of the equation.

$x^{2} + 4 x + 17 - 8 + 2 x = 0$

Simplify $x^{2} + 4 x + 17 - 8 + 2 x$ .

Add $4 x$ and $2 x$ .

$x^{2} + 6 x + 17 - 8 = 0$

Subtract $8$ from $17$ .

$x^{2} + 6 x + 9 = 0$

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 $a = 1$ , $b = 6$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 9$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{- 6 \pm \sqrt{36 - 36}}{2 \cdot 1}$

Subtract $36$ from $36$ .

$x = \frac{- 6 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{- 6 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$x = \frac{- 6 \pm 0}{2 \cdot 1}$

$- 6$ plus or minus $0$ is $- 6$ .

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

Multiply $2$ by $1$ .

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

Divide $- 6$ by $2$ .

$x = - 3$

Step 5

The final answer is the combination of both solutions.

$x = - 3$ Double roots