Solve Using the Quadratic Formula h^2-17h=-60

$h^{2} - 17 h = - 60$

Step 1

Add $60$ to both sides of the equation.

$h^{2} - 17 h + 60 = 0$

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 $a = 1$ , $b = - 17$ , and $c = 60$ into the quadratic formula and solve for $h$ .

$\frac{17 \pm \sqrt{{(- 17)}^{2} - 4 \cdot (1 \cdot 60)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 17$ to the power of $2$ .

$h = \frac{17 \pm \sqrt{289 - 4 \cdot 1 \cdot 60}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 60$ .

Multiply $- 4$ by $1$ .

$h = \frac{17 \pm \sqrt{289 - 4 \cdot 60}}{2 \cdot 1}$

Multiply $- 4$ by $60$ .

$h = \frac{17 \pm \sqrt{289 - 240}}{2 \cdot 1}$

Subtract $240$ from $289$ .

$h = \frac{17 \pm \sqrt{49}}{2 \cdot 1}$

Rewrite $49$ as $7^{2}$ .

$h = \frac{17 \pm \sqrt{7^{2}}}{2 \cdot 1}$

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

$h = \frac{17 \pm 7}{2 \cdot 1}$

Multiply $2$ by $1$ .

$h = \frac{17 \pm 7}{2}$

Step 5

The final answer is the combination of both solutions.

$h = 12, 5$