Solve Using the Quadratic Formula 2h^2=h^2+2h+17

$2 h^{2} = h^{2} + 2 h + 17$

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

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

Subtract $h^{2}$ from both sides of the equation.

$2 h^{2} - h^{2} = 2 h + 17$

Subtract $2 h$ from both sides of the equation.

$2 h^{2} - h^{2} - 2 h = 17$

Subtract $17$ from both sides of the equation.

$2 h^{2} - h^{2} - 2 h - 17 = 0$

Subtract $h^{2}$ from $2 h^{2}$ .

$h^{2} - 2 h - 17 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 17$ into the quadratic formula and solve for $h$ .

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

Simplify.

Simplify the numerator.

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

$h = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 17}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$h = \frac{2 \pm \sqrt{4 - 4 \cdot - 17}}{2 \cdot 1}$

Multiply $- 4$ by $- 17$ .

$h = \frac{2 \pm \sqrt{4 + 68}}{2 \cdot 1}$

Add $4$ and $68$ .

$h = \frac{2 \pm \sqrt{72}}{2 \cdot 1}$

Rewrite $72$ as $6^{2} \cdot 2$ .

Factor $36$ out of $72$ .

$h = \frac{2 \pm \sqrt{36 (2)}}{2 \cdot 1}$

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

$h = \frac{2 \pm \sqrt{6^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$h = \frac{2 \pm 6 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$h = \frac{2 \pm 6 \sqrt{2}}{2}$

Simplify $\frac{2 \pm 6 \sqrt{2}}{2}$ .

$h = 1 \pm 3 \sqrt{2}$

The result can be shown in multiple forms.

Exact Form:

$h = 1 \pm 3 \sqrt{2}$

Decimal Form:

$h = 5.24264068 \dots, - 3.24264068 \dots$