Solve Using the Quadratic Formula 2y^2-3y-2=0

$2 y^{2} - 3 y - 2 = 0$

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 $a = 2$ , $b = - 3$ , and $c = - 2$ into the quadratic formula and solve for $y$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 2)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$y = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 2}}{2 \cdot 2}$

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

Multiply $- 4$ by $2$ .

$y = \frac{3 \pm \sqrt{9 - 8 \cdot - 2}}{2 \cdot 2}$

Multiply $- 8$ by $- 2$ .

$y = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 2}$

Add $9$ and $16$ .

$y = \frac{3 \pm \sqrt{25}}{2 \cdot 2}$

Rewrite $25$ as $5^{2}$ .

$y = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 2}$

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

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

Multiply $2$ by $2$ .

$y = \frac{3 \pm 5}{4}$

Step 4

The final answer is the combination of both solutions.

$y = 2, - \frac{1}{2}$