Solve Using the Quadratic Formula 2y^2-6y-8=0

$2 y^{2} - 6 y - 8 = 0$

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 $a = 2$ , $b = - 6$ , and $c = - 8$ into the quadratic formula and solve for $y$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (2 \cdot - 8)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot - 8}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 8$ .

Multiply $- 4$ by $2$ .

$y = \frac{6 \pm \sqrt{36 - 8 \cdot - 8}}{2 \cdot 2}$

Multiply $- 8$ by $- 8$ .

$y = \frac{6 \pm \sqrt{36 + 64}}{2 \cdot 2}$

Add $36$ and $64$ .

$y = \frac{6 \pm \sqrt{100}}{2 \cdot 2}$

Rewrite $100$ as $10^{2}$ .

$y = \frac{6 \pm \sqrt{10^{2}}}{2 \cdot 2}$

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

$y = \frac{6 \pm 10}{2 \cdot 2}$

Multiply $2$ by $2$ .

$y = \frac{6 \pm 10}{4}$

Simplify $\frac{6 \pm 10}{4}$ .

$y = \frac{3 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$y = 4, - 1$