Solve (y+2)^2=(3y-1)^2

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

Solve for y

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$y=-\frac{1}{4}=-0.25$$

Steps Using Factoring By Grouping

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-1\right)^{2}.

$$y^{2}+4y+4=9y^{2}-6y+1$$

Subtract 9y^{2} from both sides.

$$y^{2}+4y+4-9y^{2}=-6y+1$$

Combine y^{2} and -9y^{2} to get -8y^{2}.

$$-8y^{2}+4y+4=-6y+1$$

Add 6y to both sides.

$$-8y^{2}+4y+4+6y=1$$

Combine 4y and 6y to get 10y.

$$-8y^{2}+10y+4=1$$

Subtract 1 from both sides.

$$-8y^{2}+10y+4-1=0$$

Subtract 1 from 4 to get 3.

$$-8y^{2}+10y+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -8y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=10 ab=-8\times 3=-24$$

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. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 10.

$$a=12 b=-2$$

Rewrite -8y^{2}+10y+3 as \left(-8y^{2}+12y\right)+\left(-2y+3\right).

$$\left(-8y^{2}+12y\right)+\left(-2y+3\right)$$

Factor out -4y in the first and -1 in the second group.

$$-4y\left(2y-3\right)-\left(2y-3\right)$$

Factor out common term 2y-3 by using distributive property.

$$\left(2y-3\right)\left(-4y-1\right)$$

To find equation solutions, solve 2y-3=0 and -4y-1=0.

$$y=\frac{3}{2} y=-\frac{1}{4}$$

Steps Using the Quadratic Formula

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-1\right)^{2}.

$$y^{2}+4y+4=9y^{2}-6y+1$$

Subtract 9y^{2} from both sides.

$$y^{2}+4y+4-9y^{2}=-6y+1$$

Combine y^{2} and -9y^{2} to get -8y^{2}.

$$-8y^{2}+4y+4=-6y+1$$

Add 6y to both sides.

$$-8y^{2}+4y+4+6y=1$$

Combine 4y and 6y to get 10y.

$$-8y^{2}+10y+4=1$$

Subtract 1 from both sides.

$$-8y^{2}+10y+4-1=0$$

Subtract 1 from 4 to get 3.

$$-8y^{2}+10y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-10±\sqrt{10^{2}-4\left(-8\right)\times 3}}{2\left(-8\right)}$$

Square 10.

$$y=\frac{-10±\sqrt{100-4\left(-8\right)\times 3}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$y=\frac{-10±\sqrt{100+32\times 3}}{2\left(-8\right)}$$

Multiply 32 times 3.

$$y=\frac{-10±\sqrt{100+96}}{2\left(-8\right)}$$

Add 100 to 96.

$$y=\frac{-10±\sqrt{196}}{2\left(-8\right)}$$

Take the square root of 196.

$$y=\frac{-10±14}{2\left(-8\right)}$$

Multiply 2 times -8.

$$y=\frac{-10±14}{-16}$$

Now solve the equation y=\frac{-10±14}{-16} when ± is plus. Add -10 to 14.

$$y=\frac{4}{-16}$$

Reduce the fraction \frac{4}{-16} to lowest terms by extracting and canceling out 4.

$$y=-\frac{1}{4}$$

Now solve the equation y=\frac{-10±14}{-16} when ± is minus. Subtract 14 from -10.

$$y=-\frac{24}{-16}$$

Reduce the fraction \frac{-24}{-16} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$y=-\frac{1}{4} y=\frac{3}{2}$$

Steps for Completing the Square

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-1\right)^{2}.

$$y^{2}+4y+4=9y^{2}-6y+1$$

Subtract 9y^{2} from both sides.

$$y^{2}+4y+4-9y^{2}=-6y+1$$

Combine y^{2} and -9y^{2} to get -8y^{2}.

$$-8y^{2}+4y+4=-6y+1$$

Add 6y to both sides.

$$-8y^{2}+4y+4+6y=1$$

Combine 4y and 6y to get 10y.

$$-8y^{2}+10y+4=1$$

Subtract 4 from both sides.

$$-8y^{2}+10y=1-4$$

Subtract 4 from 1 to get -3.

$$-8y^{2}+10y=-3$$

Divide both sides by -8.

$$\frac{-8y^{2}+10y}{-8}=-\frac{3}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$y^{2}+\frac{10}{-8}y=-\frac{3}{-8}$$

Reduce the fraction \frac{10}{-8} to lowest terms by extracting and canceling out 2.

$$y^{2}-\frac{5}{4}y=-\frac{3}{-8}$$

Divide -3 by -8.

$$y^{2}-\frac{5}{4}y=\frac{3}{8}$$

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{5}{4}y+\left(-\frac{5}{8}\right)^{2}=\frac{3}{8}+\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{5}{4}y+\frac{25}{64}=\frac{3}{8}+\frac{25}{64}$$

Add \frac{3}{8} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{5}{4}y+\frac{25}{64}=\frac{49}{64}$$

Factor y^{2}-\frac{5}{4}y+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{5}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{5}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$y-\frac{5}{8}=\frac{7}{8} y-\frac{5}{8}=-\frac{7}{8}$$

Add \frac{5}{8} to both sides of the equation.

$$y=\frac{3}{2} y=-\frac{1}{4}$$