Solve (y-4)^2=(y-4)(2y-1)

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

Solve for y

$$y=-3$$

$$y=4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$y^{2}-8y+16=\left(y-4\right)\left(2y-1\right)$$

Use the distributive property to multiply y-4 by 2y-1 and combine like terms.

$$y^{2}-8y+16=2y^{2}-9y+4$$

Subtract 2y^{2} from both sides.

$$y^{2}-8y+16-2y^{2}=-9y+4$$

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

$$-y^{2}-8y+16=-9y+4$$

Add 9y to both sides.

$$-y^{2}-8y+16+9y=4$$

Combine -8y and 9y to get y.

$$-y^{2}+y+16=4$$

Subtract 4 from both sides.

$$-y^{2}+y+16-4=0$$

Subtract 4 from 16 to get 12.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+12. To find a and b, set up a system to be solved.

$$a+b=1 ab=-12=-12$$

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 -12.

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

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -y^{2}+y+12 as \left(-y^{2}+4y\right)+\left(-3y+12\right).

$$\left(-y^{2}+4y\right)+\left(-3y+12\right)$$

Factor out -y in the first and -3 in the second group.

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

Factor out common term y-4 by using distributive property.

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

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

$$y=4 y=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$y^{2}-8y+16=\left(y-4\right)\left(2y-1\right)$$

Use the distributive property to multiply y-4 by 2y-1 and combine like terms.

$$y^{2}-8y+16=2y^{2}-9y+4$$

Subtract 2y^{2} from both sides.

$$y^{2}-8y+16-2y^{2}=-9y+4$$

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

$$-y^{2}-8y+16=-9y+4$$

Add 9y to both sides.

$$-y^{2}-8y+16+9y=4$$

Combine -8y and 9y to get y.

$$-y^{2}+y+16=4$$

Subtract 4 from both sides.

$$-y^{2}+y+16-4=0$$

Subtract 4 from 16 to get 12.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-1±\sqrt{1+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$y=\frac{-1±\sqrt{1+48}}{2\left(-1\right)}$$

Add 1 to 48.

$$y=\frac{-1±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$y=\frac{-1±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-1±7}{-2}$$

Now solve the equation y=\frac{-1±7}{-2} when ± is plus. Add -1 to 7.

$$y=\frac{6}{-2}$$

Divide 6 by -2.

$$y=-3$$

Now solve the equation y=\frac{-1±7}{-2} when ± is minus. Subtract 7 from -1.

$$y=-\frac{8}{-2}$$

Divide -8 by -2.

$$y=4$$

The equation is now solved.

$$y=-3 y=4$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$y^{2}-8y+16=\left(y-4\right)\left(2y-1\right)$$

Use the distributive property to multiply y-4 by 2y-1 and combine like terms.

$$y^{2}-8y+16=2y^{2}-9y+4$$

Subtract 2y^{2} from both sides.

$$y^{2}-8y+16-2y^{2}=-9y+4$$

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

$$-y^{2}-8y+16=-9y+4$$

Add 9y to both sides.

$$-y^{2}-8y+16+9y=4$$

Combine -8y and 9y to get y.

$$-y^{2}+y+16=4$$

Subtract 16 from both sides.

$$-y^{2}+y=4-16$$

Subtract 16 from 4 to get -12.

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

Divide both sides by -1.

$$\frac{-y^{2}+y}{-1}=-\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{1}{-1}y=-\frac{12}{-1}$$

Divide 1 by -1.

$$y^{2}-y=-\frac{12}{-1}$$

Divide -12 by -1.

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

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

$$y^{2}-y+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-y+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$y^{2}-y+\frac{1}{4}=\frac{49}{4}$$

Factor y^{2}-y+\frac{1}{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}.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$y-\frac{1}{2}=\frac{7}{2} y-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$y=4 y=-3$$