Solve Using the Quadratic Formula y^2-7y-18=0

$y^{2} - 7 y - 18 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 7$ , and $c = - 18$ into the quadratic formula and solve for $y$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (1 \cdot - 18)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$y = \frac{7 \pm \sqrt{49 - 4 \cdot 1 \cdot - 18}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 18$ .

Multiply $- 4$ by $1$ .

$y = \frac{7 \pm \sqrt{49 - 4 \cdot - 18}}{2 \cdot 1}$

Multiply $- 4$ by $- 18$ .

$y = \frac{7 \pm \sqrt{49 + 72}}{2 \cdot 1}$

Add $49$ and $72$ .

$y = \frac{7 \pm \sqrt{121}}{2 \cdot 1}$

Rewrite $121$ as $11^{2}$ .

$y = \frac{7 \pm \sqrt{11^{2}}}{2 \cdot 1}$

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

$y = \frac{7 \pm 11}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{7 \pm 11}{2}$

Step 4

The final answer is the combination of both solutions.

$y = 9, - 2$