Solve 2(y+1)^2=y(y+1)

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

Solve for y

$$y=-2$$

$$y=-1$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$2\left(y^{2}+2y+1\right)=y\left(y+1\right)$$

Use the distributive property to multiply 2 by y^{2}+2y+1.

$$2y^{2}+4y+2=y\left(y+1\right)$$

Use the distributive property to multiply y by y+1.

$$2y^{2}+4y+2=y^{2}+y$$

Subtract y^{2} from both sides.

$$2y^{2}+4y+2-y^{2}=y$$

Combine 2y^{2} and -y^{2} to get y^{2}.

$$y^{2}+4y+2=y$$

Subtract y from both sides.

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

Combine 4y and -y to get 3y.

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

To solve the equation, factor y^{2}+3y+2 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y+1\right)\left(y+2\right)$$

To find equation solutions, solve y+1=0 and y+2=0.

$$y=-1 y=-2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$2\left(y^{2}+2y+1\right)=y\left(y+1\right)$$

Use the distributive property to multiply 2 by y^{2}+2y+1.

$$2y^{2}+4y+2=y\left(y+1\right)$$

Use the distributive property to multiply y by y+1.

$$2y^{2}+4y+2=y^{2}+y$$

Subtract y^{2} from both sides.

$$2y^{2}+4y+2-y^{2}=y$$

Combine 2y^{2} and -y^{2} to get y^{2}.

$$y^{2}+4y+2=y$$

Subtract y from both sides.

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

Combine 4y and -y to get 3y.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+2. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite y^{2}+3y+2 as \left(y^{2}+y\right)+\left(2y+2\right).

$$\left(y^{2}+y\right)+\left(2y+2\right)$$

Factor out y in the first and 2 in the second group.

$$y\left(y+1\right)+2\left(y+1\right)$$

Factor out common term y+1 by using distributive property.

$$\left(y+1\right)\left(y+2\right)$$

To find equation solutions, solve y+1=0 and y+2=0.

$$y=-1 y=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$2\left(y^{2}+2y+1\right)=y\left(y+1\right)$$

Use the distributive property to multiply 2 by y^{2}+2y+1.

$$2y^{2}+4y+2=y\left(y+1\right)$$

Use the distributive property to multiply y by y+1.

$$2y^{2}+4y+2=y^{2}+y$$

Subtract y^{2} from both sides.

$$2y^{2}+4y+2-y^{2}=y$$

Combine 2y^{2} and -y^{2} to get y^{2}.

$$y^{2}+4y+2=y$$

Subtract y from both sides.

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

Combine 4y and -y to get 3y.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-3±\sqrt{3^{2}-4\times 2}}{2}$$

Square 3.

$$y=\frac{-3±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

$$y=\frac{-3±\sqrt{9-8}}{2}$$

Add 9 to -8.

$$y=\frac{-3±\sqrt{1}}{2}$$

Take the square root of 1.

$$y=\frac{-3±1}{2}$$

Now solve the equation y=\frac{-3±1}{2} when ± is plus. Add -3 to 1.

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

Divide -2 by 2.

$$y=-1$$

Now solve the equation y=\frac{-3±1}{2} when ± is minus. Subtract 1 from -3.

$$y=-\frac{4}{2}$$

Divide -4 by 2.

$$y=-2$$

The equation is now solved.

$$y=-1 y=-2$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$2\left(y^{2}+2y+1\right)=y\left(y+1\right)$$

Use the distributive property to multiply 2 by y^{2}+2y+1.

$$2y^{2}+4y+2=y\left(y+1\right)$$

Use the distributive property to multiply y by y+1.

$$2y^{2}+4y+2=y^{2}+y$$

Subtract y^{2} from both sides.

$$2y^{2}+4y+2-y^{2}=y$$

Combine 2y^{2} and -y^{2} to get y^{2}.

$$y^{2}+4y+2=y$$

Subtract y from both sides.

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

Combine 4y and -y to get 3y.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+3y+\left(\frac{3}{2}\right)^{2}=-2+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+3y+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$y^{2}+3y+\frac{9}{4}=\frac{1}{4}$$

Factor y^{2}+3y+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$y+\frac{3}{2}=\frac{1}{2} y+\frac{3}{2}=-\frac{1}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$y=-1 y=-2$$