Solve 2y^2-y+66=(y-6)^2

$$2 y ^ { 2 } - y + 66 = ( y - 6 ) ^ { 2 }$$

Solve for y

$$y=-6$$

$$y=-5$$

Steps Using Factoring

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

$$2y^{2}-y+66=y^{2}-12y+36$$

Subtract y^{2} from both sides.

$$2y^{2}-y+66-y^{2}=-12y+36$$

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

$$y^{2}-y+66=-12y+36$$

Add 12y to both sides.

$$y^{2}-y+66+12y=36$$

Combine -y and 12y to get 11y.

$$y^{2}+11y+66=36$$

Subtract 36 from both sides.

$$y^{2}+11y+66-36=0$$

Subtract 36 from 66 to get 30.

$$y^{2}+11y+30=0$$

To solve the equation, factor y^{2}+11y+30 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.

$$a+b=11 ab=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=5 b=6$$

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

$$\left(y+5\right)\left(y+6\right)$$

To find equation solutions, solve y+5=0 and y+6=0.

$$y=-5 y=-6$$

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}.

$$2y^{2}-y+66=y^{2}-12y+36$$

Subtract y^{2} from both sides.

$$2y^{2}-y+66-y^{2}=-12y+36$$

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

$$y^{2}-y+66=-12y+36$$

Add 12y to both sides.

$$y^{2}-y+66+12y=36$$

Combine -y and 12y to get 11y.

$$y^{2}+11y+66=36$$

Subtract 36 from both sides.

$$y^{2}+11y+66-36=0$$

Subtract 36 from 66 to get 30.

$$y^{2}+11y+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+30. To find a and b, set up a system to be solved.

$$a+b=11 ab=1\times 30=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=5 b=6$$

Rewrite y^{2}+11y+30 as \left(y^{2}+5y\right)+\left(6y+30\right).

$$\left(y^{2}+5y\right)+\left(6y+30\right)$$

Factor out y in the first and 6 in the second group.

$$y\left(y+5\right)+6\left(y+5\right)$$

Factor out common term y+5 by using distributive property.

$$\left(y+5\right)\left(y+6\right)$$

To find equation solutions, solve y+5=0 and y+6=0.

$$y=-5 y=-6$$

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}.

$$2y^{2}-y+66=y^{2}-12y+36$$

Subtract y^{2} from both sides.

$$2y^{2}-y+66-y^{2}=-12y+36$$

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

$$y^{2}-y+66=-12y+36$$

Add 12y to both sides.

$$y^{2}-y+66+12y=36$$

Combine -y and 12y to get 11y.

$$y^{2}+11y+66=36$$

Subtract 36 from both sides.

$$y^{2}+11y+66-36=0$$

Subtract 36 from 66 to get 30.

$$y^{2}+11y+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 11 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-11±\sqrt{11^{2}-4\times 30}}{2}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\times 30}}{2}$$

Multiply -4 times 30.

$$y=\frac{-11±\sqrt{121-120}}{2}$$

Add 121 to -120.

$$y=\frac{-11±\sqrt{1}}{2}$$

Take the square root of 1.

$$y=\frac{-11±1}{2}$$

Now solve the equation y=\frac{-11±1}{2} when ± is plus. Add -11 to 1.

$$y=-\frac{10}{2}$$

Divide -10 by 2.

$$y=-5$$

Now solve the equation y=\frac{-11±1}{2} when ± is minus. Subtract 1 from -11.

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

Divide -12 by 2.

$$y=-6$$

The equation is now solved.

$$y=-5 y=-6$$

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}.

$$2y^{2}-y+66=y^{2}-12y+36$$

Subtract y^{2} from both sides.

$$2y^{2}-y+66-y^{2}=-12y+36$$

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

$$y^{2}-y+66=-12y+36$$

Add 12y to both sides.

$$y^{2}-y+66+12y=36$$

Combine -y and 12y to get 11y.

$$y^{2}+11y+66=36$$

Subtract 66 from both sides.

$$y^{2}+11y=36-66$$

Subtract 66 from 36 to get -30.

$$y^{2}+11y=-30$$

Divide 11, the coefficient of the x term, by 2 to get \frac{11}{2}. Then add the square of \frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+11y+\left(\frac{11}{2}\right)^{2}=-30+\left(\frac{11}{2}\right)^{2}$$

Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+11y+\frac{121}{4}=-30+\frac{121}{4}$$

Add -30 to \frac{121}{4}.

$$y^{2}+11y+\frac{121}{4}=\frac{1}{4}$$

Factor y^{2}+11y+\frac{121}{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}.

$$\left(y+\frac{11}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{11}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$y+\frac{11}{2}=\frac{1}{2} y+\frac{11}{2}=-\frac{1}{2}$$

Subtract \frac{11}{2} from both sides of the equation.

$$y=-5 y=-6$$