Solve Using the Quadratic Formula 2y^2-28-7y=y^2-4y

$2 y^{2} - 28 - 7 y = y^{2} - 4 y$

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.

$2 y^{2} - 28 - 7 y - y^{2} = - 4 y$

Add $4 y$ to both sides of the equation.

$2 y^{2} - 28 - 7 y - y^{2} + 4 y = 0$

Simplify $2 y^{2} - 28 - 7 y - y^{2} + 4 y$ .

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

$y^{2} - 28 - 7 y + 4 y = 0$

Add $- 7 y$ and $4 y$ .

$y^{2} - 28 - 3 y = 0$

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 $a = 1$ , $b = - 3$ , and $c = - 28$ into the quadratic formula and solve for $y$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot - 28)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$y = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 28}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 28$ .

Multiply $- 4$ by $1$ .

$y = \frac{3 \pm \sqrt{9 - 4 \cdot - 28}}{2 \cdot 1}$

Multiply $- 4$ by $- 28$ .

$y = \frac{3 \pm \sqrt{9 + 112}}{2 \cdot 1}$

Add $9$ and $112$ .

$y = \frac{3 \pm \sqrt{121}}{2 \cdot 1}$

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

$y = \frac{3 \pm \sqrt{11^{2}}}{2 \cdot 1}$

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

$y = \frac{3 \pm 11}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{3 \pm 11}{2}$

Step 5

The final answer is the combination of both solutions.

$y = 7, - 4$