Solve Using the Quadratic Formula 2y^2-15-5y=y^2-3y

$2 y^{2} - 15 - 5 y = y^{2} - 3 y$

Step 1

Move all terms to the left side of the equation and simplify.

Move all the expressions to the left side of the equation.

Subtract $y^{2}$ from both sides of the equation.

$2 y^{2} - 15 - 5 y - y^{2} = - 3 y$

Add $3 y$ to both sides of the equation.

$2 y^{2} - 15 - 5 y - y^{2} + 3 y = 0$

Simplify $2 y^{2} - 15 - 5 y - y^{2} + 3 y$ .

Subtract $y^{2}$ from $2 y^{2}$ .

$y^{2} - 15 - 5 y + 3 y = 0$

Add $- 5 y$ and $3 y$ .

$y^{2} - 15 - 2 y = 0$

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 $a = 1$ , $b = - 2$ , and $c = - 15$ into the quadratic formula and solve for $y$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 15)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 15$ .

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$y = \frac{2 \pm \sqrt{4 + 60}}{2 \cdot 1}$

Add $4$ and $60$ .

$y = \frac{2 \pm \sqrt{64}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$y = \frac{2 \pm \sqrt{8^{2}}}{2 \cdot 1}$

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

$y = \frac{2 \pm 8}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 8}{2}$

Simplify $\frac{2 \pm 8}{2}$ .

$y = 1 \pm 4$

Step 5

The final answer is the combination of both solutions.

$y = 5, - 3$