Solve left(3y-2right)^2=y^2

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

Solve for y

$$y=\frac{1}{2}=0.5$$

$$y=1$$

Steps Using Factoring By Grouping

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

$$9y^{2}-12y+4=y^{2}$$

Subtract y^{2} from both sides.

$$9y^{2}-12y+4-y^{2}=0$$

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

$$8y^{2}-12y+4=0$$

Divide both sides by 4.

$$2y^{2}-3y+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2y^{2}+ay+by+1. To find a and b, set up a system to be solved.

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

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

$$a=-2 b=-1$$

Rewrite 2y^{2}-3y+1 as \left(2y^{2}-2y\right)+\left(-y+1\right).

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

Factor out 2y in the first and -1 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

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

$$9y^{2}-12y+4=y^{2}$$

Subtract y^{2} from both sides.

$$9y^{2}-12y+4-y^{2}=0$$

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

$$8y^{2}-12y+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -12 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 8\times 4}}{2\times 8}$$

Square -12.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 8\times 4}}{2\times 8}$$

Multiply -4 times 8.

$$y=\frac{-\left(-12\right)±\sqrt{144-32\times 4}}{2\times 8}$$

Multiply -32 times 4.

$$y=\frac{-\left(-12\right)±\sqrt{144-128}}{2\times 8}$$

Add 144 to -128.

$$y=\frac{-\left(-12\right)±\sqrt{16}}{2\times 8}$$

Take the square root of 16.

$$y=\frac{-\left(-12\right)±4}{2\times 8}$$

The opposite of -12 is 12.

$$y=\frac{12±4}{2\times 8}$$

Multiply 2 times 8.

$$y=\frac{12±4}{16}$$

Now solve the equation y=\frac{12±4}{16} when ± is plus. Add 12 to 4.

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

Divide 16 by 16.

$$y=1$$

Now solve the equation y=\frac{12±4}{16} when ± is minus. Subtract 4 from 12.

$$y=\frac{8}{16}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

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

$$9y^{2}-12y+4=y^{2}$$

Subtract y^{2} from both sides.

$$9y^{2}-12y+4-y^{2}=0$$

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

$$8y^{2}-12y+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$8y^{2}-12y=-4$$

Divide both sides by 8.

$$\frac{8y^{2}-12y}{8}=-\frac{4}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$y^{2}+\left(-\frac{12}{8}\right)y=-\frac{4}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{3}{2}y+\frac{9}{16}=-\frac{1}{2}+\frac{9}{16}$$

Add -\frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{3}{2}y+\frac{9}{16}=\frac{1}{16}$$

Factor y^{2}-\frac{3}{2}y+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{3}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

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

Add \frac{3}{4} to both sides of the equation.

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