Solve Using the Quadratic Formula 2y^2-3y-6=0

$2 y^{2} - 3 y - 6 = 0$

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 $c = - 6$ into the quadratic formula and solve for $y$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 6)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 6}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 6$ .

Multiply $- 4$ by $2$ .

$y = \frac{3 \pm \sqrt{9 - 8 \cdot - 6}}{2 \cdot 2}$

Multiply $- 8$ by $- 6$ .

$y = \frac{3 \pm \sqrt{9 + 48}}{2 \cdot 2}$

Add $9$ and $48$ .

$y = \frac{3 \pm \sqrt{57}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$y = \frac{3 \pm \sqrt{57}}{4}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$y = \frac{3 \pm \sqrt{57}}{4}$

Decimal Form:

$y = 2.63745860 \dots, - 1.13745860 \dots$