Solve (2y-1)^2-2(2y-1)-3=0

$$( 2 y - 1 ) ^ { 2 } - 2 ( 2 y - 1 ) - 3 = 0$$

Solve for y

$$y=0$$

$$y=2$$

Steps Using Factoring

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

$$4y^{2}-4y+1-2\left(2y-1\right)-3=0$$

Use the distributive property to multiply -2 by 2y-1.

$$4y^{2}-4y+1-4y+2-3=0$$

Combine -4y and -4y to get -8y.

$$4y^{2}-8y+1+2-3=0$$

Add 1 and 2 to get 3.

$$4y^{2}-8y+3-3=0$$

Subtract 3 from 3 to get 0.

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

Factor out y.

$$y\left(4y-8\right)=0$$

To find equation solutions, solve y=0 and 4y-8=0.

$$y=0 y=2$$

Steps Using the Quadratic Formula

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

$$4y^{2}-4y+1-2\left(2y-1\right)-3=0$$

Use the distributive property to multiply -2 by 2y-1.

$$4y^{2}-4y+1-4y+2-3=0$$

Combine -4y and -4y to get -8y.

$$4y^{2}-8y+1+2-3=0$$

Add 1 and 2 to get 3.

$$4y^{2}-8y+3-3=0$$

Subtract 3 from 3 to get 0.

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

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

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

Take the square root of \left(-8\right)^{2}.

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

The opposite of -8 is 8.

$$y=\frac{8±8}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{8±8}{8}$$

Now solve the equation y=\frac{8±8}{8} when ± is plus. Add 8 to 8.

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

Divide 16 by 8.

$$y=2$$

Now solve the equation y=\frac{8±8}{8} when ± is minus. Subtract 8 from 8.

$$y=\frac{0}{8}$$

Divide 0 by 8.

$$y=0$$

The equation is now solved.

$$y=2 y=0$$

Steps for Completing the Square

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

$$4y^{2}-4y+1-2\left(2y-1\right)-3=0$$

Use the distributive property to multiply -2 by 2y-1.

$$4y^{2}-4y+1-4y+2-3=0$$

Combine -4y and -4y to get -8y.

$$4y^{2}-8y+1+2-3=0$$

Add 1 and 2 to get 3.

$$4y^{2}-8y+3-3=0$$

Subtract 3 from 3 to get 0.

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

Divide both sides by 4.

$$\frac{4y^{2}-8y}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\left(-\frac{8}{4}\right)y=\frac{0}{4}$$

Divide -8 by 4.

$$y^{2}-2y=\frac{0}{4}$$

Divide 0 by 4.

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

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.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 1 to both sides of the equation.

$$y=2 y=0$$