Solve (y+3)^2=2y(y+y)

$$( y + 3 ) ^ { 2 } = 2 y ( y + y )$$

Solve for y

$$y=3$$

$$y=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$y^{2}+6y+9=2y\left(y+y\right)$$

Combine y and y to get 2y.

$$y^{2}+6y+9=2y\times 2y$$

Multiply 2 and 2 to get 4.

$$y^{2}+6y+9=4yy$$

Multiply y and y to get y^{2}.

$$y^{2}+6y+9=4y^{2}$$

Subtract 4y^{2} from both sides.

$$y^{2}+6y+9-4y^{2}=0$$

Combine y^{2} and -4y^{2} to get -3y^{2}.

$$-3y^{2}+6y+9=0$$

Divide both sides by 3.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+3. To find a and b, set up a system to be solved.

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

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

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

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

Factor out -y in the first and -1 in the second group.

$$-y\left(y-3\right)-\left(y-3\right)$$

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(-y-1\right)$$

To find equation solutions, solve y-3=0 and -y-1=0.

$$y=3 y=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$y^{2}+6y+9=2y\left(y+y\right)$$

Combine y and y to get 2y.

$$y^{2}+6y+9=2y\times 2y$$

Multiply 2 and 2 to get 4.

$$y^{2}+6y+9=4yy$$

Multiply y and y to get y^{2}.

$$y^{2}+6y+9=4y^{2}$$

Subtract 4y^{2} from both sides.

$$y^{2}+6y+9-4y^{2}=0$$

Combine y^{2} and -4y^{2} to get -3y^{2}.

$$-3y^{2}+6y+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Square 6.

$$y=\frac{-6±\sqrt{36-4\left(-3\right)\times 9}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$y=\frac{-6±\sqrt{36+12\times 9}}{2\left(-3\right)}$$

Multiply 12 times 9.

$$y=\frac{-6±\sqrt{36+108}}{2\left(-3\right)}$$

Add 36 to 108.

$$y=\frac{-6±\sqrt{144}}{2\left(-3\right)}$$

Take the square root of 144.

$$y=\frac{-6±12}{2\left(-3\right)}$$

Multiply 2 times -3.

$$y=\frac{-6±12}{-6}$$

Now solve the equation y=\frac{-6±12}{-6} when ± is plus. Add -6 to 12.

$$y=\frac{6}{-6}$$

Divide 6 by -6.

$$y=-1$$

Now solve the equation y=\frac{-6±12}{-6} when ± is minus. Subtract 12 from -6.

$$y=-\frac{18}{-6}$$

Divide -18 by -6.

$$y=3$$

The equation is now solved.

$$y=-1 y=3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$y^{2}+6y+9=2y\left(y+y\right)$$

Combine y and y to get 2y.

$$y^{2}+6y+9=2y\times 2y$$

Multiply 2 and 2 to get 4.

$$y^{2}+6y+9=4yy$$

Multiply y and y to get y^{2}.

$$y^{2}+6y+9=4y^{2}$$

Subtract 4y^{2} from both sides.

$$y^{2}+6y+9-4y^{2}=0$$

Combine y^{2} and -4y^{2} to get -3y^{2}.

$$-3y^{2}+6y+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-3y^{2}+6y=-9$$

Divide both sides by -3.

$$\frac{-3y^{2}+6y}{-3}=-\frac{9}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$y^{2}+\frac{6}{-3}y=-\frac{9}{-3}$$

Divide 6 by -3.

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

Divide -9 by -3.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-2y+1=3+1$$

Add 3 to 1.

$$y^{2}-2y+1=4$$

Factor y^{2}-2y+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

$$y=3 y=-1$$