Solve Using the Quadratic Formula x^2+2x+1=17

$x^{2} + 2 x + 1 = 17$

Step 1

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

Subtract $17$ from both sides of the equation.

$x^{2} + 2 x + 1 - 17 = 0$

Subtract $17$ from $1$ .

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

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 $a = 1$ , $b = 2$ , and $c = - 16$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 16$ .

$x = \frac{- 2 \pm \sqrt{4 + 64}}{2 \cdot 1}$

Add $4$ and $64$ .

$x = \frac{- 2 \pm \sqrt{68}}{2 \cdot 1}$

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x = \frac{- 2 \pm \sqrt{4 (17)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 17}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{17}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{17}}{2}$

Simplify $\frac{- 2 \pm 2 \sqrt{17}}{2}$ .

$x = - 1 \pm \sqrt{17}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - 1 \pm \sqrt{17}$

Decimal Form:

$x = 3.12310562 \dots, - 5.12310562 \dots$