Solve for y (2y+1)(y-1)=(y+5(2y-5))

$(2 y + 1) (y - 1) = (y + 5 (2 y - 5))$

Step 1

Simplify $(2 y + 1) (y - 1)$ .

Rewrite.

$0 + 0 + (2 y + 1) (y - 1) = y + 5 (2 y - 5)$

Simplify by adding zeros.

$(2 y + 1) (y - 1) = y + 5 (2 y - 5)$

Expand $(2 y + 1) (y - 1)$ using the FOIL Method.

Apply the distributive property.

$2 y (y - 1) + 1 (y - 1) = y + 5 (2 y - 5)$

Apply the distributive property.

$2 y \cdot y + 2 y \cdot - 1 + 1 (y - 1) = y + 5 (2 y - 5)$

Apply the distributive property.

$2 y \cdot y + 2 y \cdot - 1 + 1 y + 1 \cdot - 1 = y + 5 (2 y - 5)$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$2 (y \cdot y) + 2 y \cdot - 1 + 1 y + 1 \cdot - 1 = y + 5 (2 y - 5)$

Multiply $y$ by $y$ .

$2 y^{2} + 2 y \cdot - 1 + 1 y + 1 \cdot - 1 = y + 5 (2 y - 5)$

Multiply $- 1$ by $2$ .

$2 y^{2} - 2 y + 1 y + 1 \cdot - 1 = y + 5 (2 y - 5)$

Multiply $y$ by $1$ .

$2 y^{2} - 2 y + y + 1 \cdot - 1 = y + 5 (2 y - 5)$

Multiply $- 1$ by $1$ .

$2 y^{2} - 2 y + y - 1 = y + 5 (2 y - 5)$

Add $- 2 y$ and $y$ .

$2 y^{2} - y - 1 = y + 5 (2 y - 5)$

Step 2

Simplify $y + 5 (2 y - 5)$ .

Simplify each term.

Apply the distributive property.

$2 y^{2} - y - 1 = y + 5 (2 y) + 5 \cdot - 5$

Multiply $2$ by $5$ .

$2 y^{2} - y - 1 = y + 10 y + 5 \cdot - 5$

Multiply $5$ by $- 5$ .

$2 y^{2} - y - 1 = y + 10 y - 25$

Add $y$ and $10 y$ .

$2 y^{2} - y - 1 = 11 y - 25$

Step 3

Move all terms containing $y$ to the left side of the equation.

Subtract $11 y$ from both sides of the equation.

$2 y^{2} - y - 1 - 11 y = - 25$

Subtract $11 y$ from $- y$ .

$2 y^{2} - 12 y - 1 = - 25$

Step 4

Add $25$ to both sides of the equation.

$2 y^{2} - 12 y - 1 + 25 = 0$

Step 5

Add $- 1$ and $25$ .

$2 y^{2} - 12 y + 24 = 0$

Step 6

Factor $2$ out of $2 y^{2} - 12 y + 24$ .

Factor $2$ out of $2 y^{2}$ .

$2 (y^{2}) - 12 y + 24 = 0$

Factor $2$ out of $- 12 y$ .

$2 (y^{2}) + 2 (- 6 y) + 24 = 0$

Factor $2$ out of $24$ .

$2 y^{2} + 2 (- 6 y) + 2 \cdot 12 = 0$

Factor $2$ out of $2 y^{2} + 2 (- 6 y)$ .

$2 (y^{2} - 6 y) + 2 \cdot 12 = 0$

Factor $2$ out of $2 (y^{2} - 6 y) + 2 \cdot 12$ .

$2 (y^{2} - 6 y + 12) = 0$

Step 7

Divide each term in $2 (y^{2} - 6 y + 12) = 0$ by $2$ and simplify.

Divide each term in $2 (y^{2} - 6 y + 12) = 0$ by $2$ .

$\frac{2 (y^{2} - 6 y + 12)}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (y^{2} - 6 y + 12)}{2} = \frac{0}{2}$

Divide $y^{2} - 6 y + 12$ by $1$ .

$y^{2} - 6 y + 12 = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$y^{2} - 6 y + 12 = 0$

Step 8

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 9

Substitute the values $a = 1$ , $b = - 6$ , and $c = 12$ into the quadratic formula and solve for $y$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 12)}}{2 \cdot 1}$

Step 10

Simplify.

Simplify the numerator.

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

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 12$ .

Multiply $- 4$ by $1$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 12}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$y = \frac{6 \pm \sqrt{36 - 48}}{2 \cdot 1}$

Subtract $48$ from $36$ .

$y = \frac{6 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{6 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{6 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$y = \frac{6 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$y = \frac{6 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{6 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$y = \frac{6 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{6 \pm 2 i \sqrt{3}}{2}$

Simplify $\frac{6 \pm 2 i \sqrt{3}}{2}$ .

$y = 3 \pm i \sqrt{3}$

Step 11

The final answer is the combination of both solutions.

$y = 3 + i \sqrt{3}, 3 - i \sqrt{3}$

Solve for y (2y+1)(y-1)=(y+5(2y-5))

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