Solve (y-6)^2-3(y-6)+2=0

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

Solve for y

$$y=7$$

$$y=8$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-6\right)^{2}.

$$y^{2}-12y+36-3\left(y-6\right)+2=0$$

Use the distributive property to multiply -3 by y-6.

$$y^{2}-12y+36-3y+18+2=0$$

Combine -12y and -3y to get -15y.

$$y^{2}-15y+36+18+2=0$$

Add 36 and 18 to get 54.

$$y^{2}-15y+54+2=0$$

Add 54 and 2 to get 56.

$$y^{2}-15y+56=0$$

To solve the equation, factor y^{2}-15y+56 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-15 ab=56$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 56.

$$-1,-56 -2,-28 -4,-14 -7,-8$$

Calculate the sum for each pair.

$$-1-56=-57 -2-28=-30 -4-14=-18 -7-8=-15$$

The solution is the pair that gives sum -15.

$$a=-8 b=-7$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-8\right)\left(y-7\right)$$

To find equation solutions, solve y-8=0 and y-7=0.

$$y=8 y=7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-6\right)^{2}.

$$y^{2}-12y+36-3\left(y-6\right)+2=0$$

Use the distributive property to multiply -3 by y-6.

$$y^{2}-12y+36-3y+18+2=0$$

Combine -12y and -3y to get -15y.

$$y^{2}-15y+36+18+2=0$$

Add 36 and 18 to get 54.

$$y^{2}-15y+54+2=0$$

Add 54 and 2 to get 56.

$$y^{2}-15y+56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+56. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 56=56$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 56.

$$-1,-56 -2,-28 -4,-14 -7,-8$$

Calculate the sum for each pair.

$$-1-56=-57 -2-28=-30 -4-14=-18 -7-8=-15$$

The solution is the pair that gives sum -15.

$$a=-8 b=-7$$

Rewrite y^{2}-15y+56 as \left(y^{2}-8y\right)+\left(-7y+56\right).

$$\left(y^{2}-8y\right)+\left(-7y+56\right)$$

Factor out y in the first and -7 in the second group.

$$y\left(y-8\right)-7\left(y-8\right)$$

Factor out common term y-8 by using distributive property.

$$\left(y-8\right)\left(y-7\right)$$

To find equation solutions, solve y-8=0 and y-7=0.

$$y=8 y=7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-6\right)^{2}.

$$y^{2}-12y+36-3\left(y-6\right)+2=0$$

Use the distributive property to multiply -3 by y-6.

$$y^{2}-12y+36-3y+18+2=0$$

Combine -12y and -3y to get -15y.

$$y^{2}-15y+36+18+2=0$$

Add 36 and 18 to get 54.

$$y^{2}-15y+54+2=0$$

Add 54 and 2 to get 56.

$$y^{2}-15y+56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -15 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 56}}{2}$$

Square -15.

$$y=\frac{-\left(-15\right)±\sqrt{225-4\times 56}}{2}$$

Multiply -4 times 56.

$$y=\frac{-\left(-15\right)±\sqrt{225-224}}{2}$$

Add 225 to -224.

$$y=\frac{-\left(-15\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$y=\frac{-\left(-15\right)±1}{2}$$

The opposite of -15 is 15.

$$y=\frac{15±1}{2}$$

Now solve the equation y=\frac{15±1}{2} when ± is plus. Add 15 to 1.

$$y=\frac{16}{2}$$

Divide 16 by 2.

$$y=8$$

Now solve the equation y=\frac{15±1}{2} when ± is minus. Subtract 1 from 15.

$$y=\frac{14}{2}$$

Divide 14 by 2.

$$y=7$$

The equation is now solved.

$$y=8 y=7$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-6\right)^{2}.

$$y^{2}-12y+36-3\left(y-6\right)+2=0$$

Use the distributive property to multiply -3 by y-6.

$$y^{2}-12y+36-3y+18+2=0$$

Combine -12y and -3y to get -15y.

$$y^{2}-15y+36+18+2=0$$

Add 36 and 18 to get 54.

$$y^{2}-15y+54+2=0$$

Add 54 and 2 to get 56.

$$y^{2}-15y+56=0$$

Subtract 56 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}-15y=-56$$

Divide -15, the coefficient of the x term, by 2 to get -\frac{15}{2}. Then add the square of -\frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-15y+\left(-\frac{15}{2}\right)^{2}=-56+\left(-\frac{15}{2}\right)^{2}$$

Square -\frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-15y+\frac{225}{4}=-56+\frac{225}{4}$$

Add -56 to \frac{225}{4}.

$$y^{2}-15y+\frac{225}{4}=\frac{1}{4}$$

Factor y^{2}-15y+\frac{225}{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}.

$$\left(y-\frac{15}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{15}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$y-\frac{15}{2}=\frac{1}{2} y-\frac{15}{2}=-\frac{1}{2}$$

Add \frac{15}{2} to both sides of the equation.

$$y=8 y=7$$