Solve (y-4)^2=2y^2-11y-12

$$( y - 4 ) ^ { 2 } = 2 y ^ { 2 } - 11 y - 12$$

Solve for y

$$y=-4$$

$$y=7$$

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}.

$$y^{2}-8y+16=2y^{2}-11y-12$$

Subtract 2y^{2} from both sides.

$$y^{2}-8y+16-2y^{2}=-11y-12$$

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

$$-y^{2}-8y+16=-11y-12$$

Add 11y to both sides.

$$-y^{2}-8y+16+11y=-12$$

Combine -8y and 11y to get 3y.

$$-y^{2}+3y+16=-12$$

Add 12 to both sides.

$$-y^{2}+3y+16+12=0$$

Add 16 and 12 to get 28.

$$-y^{2}+3y+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+28. To find a and b, set up a system to be solved.

$$a+b=3 ab=-28=-28$$

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 -28.

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

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=7 b=-4$$

Rewrite -y^{2}+3y+28 as \left(-y^{2}+7y\right)+\left(-4y+28\right).

$$\left(-y^{2}+7y\right)+\left(-4y+28\right)$$

Factor out -y in the first and -4 in the second group.

$$-y\left(y-7\right)-4\left(y-7\right)$$

Factor out common term y-7 by using distributive property.

$$\left(y-7\right)\left(-y-4\right)$$

To find equation solutions, solve y-7=0 and -y-4=0.

$$y=7 y=-4$$

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}.

$$y^{2}-8y+16=2y^{2}-11y-12$$

Subtract 2y^{2} from both sides.

$$y^{2}-8y+16-2y^{2}=-11y-12$$

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

$$-y^{2}-8y+16=-11y-12$$

Add 11y to both sides.

$$-y^{2}-8y+16+11y=-12$$

Combine -8y and 11y to get 3y.

$$-y^{2}+3y+16=-12$$

Add 12 to both sides.

$$-y^{2}+3y+16+12=0$$

Add 16 and 12 to get 28.

$$-y^{2}+3y+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\times 28}}{2\left(-1\right)}$$

Square 3.

$$y=\frac{-3±\sqrt{9-4\left(-1\right)\times 28}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-3±\sqrt{9+4\times 28}}{2\left(-1\right)}$$

Multiply 4 times 28.

$$y=\frac{-3±\sqrt{9+112}}{2\left(-1\right)}$$

Add 9 to 112.

$$y=\frac{-3±\sqrt{121}}{2\left(-1\right)}$$

Take the square root of 121.

$$y=\frac{-3±11}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-3±11}{-2}$$

Now solve the equation y=\frac{-3±11}{-2} when ± is plus. Add -3 to 11.

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

Divide 8 by -2.

$$y=-4$$

Now solve the equation y=\frac{-3±11}{-2} when ± is minus. Subtract 11 from -3.

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

Divide -14 by -2.

$$y=7$$

The equation is now solved.

$$y=-4 y=7$$

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}.

$$y^{2}-8y+16=2y^{2}-11y-12$$

Subtract 2y^{2} from both sides.

$$y^{2}-8y+16-2y^{2}=-11y-12$$

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

$$-y^{2}-8y+16=-11y-12$$

Add 11y to both sides.

$$-y^{2}-8y+16+11y=-12$$

Combine -8y and 11y to get 3y.

$$-y^{2}+3y+16=-12$$

Subtract 16 from both sides.

$$-y^{2}+3y=-12-16$$

Subtract 16 from -12 to get -28.

$$-y^{2}+3y=-28$$

Divide both sides by -1.

$$\frac{-y^{2}+3y}{-1}=-\frac{28}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{3}{-1}y=-\frac{28}{-1}$$

Divide 3 by -1.

$$y^{2}-3y=-\frac{28}{-1}$$

Divide -28 by -1.

$$y^{2}-3y=28$$

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

$$y^{2}-3y+\left(-\frac{3}{2}\right)^{2}=28+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-3y+\frac{9}{4}=28+\frac{9}{4}$$

Add 28 to \frac{9}{4}.

$$y^{2}-3y+\frac{9}{4}=\frac{121}{4}$$

Factor y^{2}-3y+\frac{9}{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}.

$$\left(y-\frac{3}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$y-\frac{3}{2}=\frac{11}{2} y-\frac{3}{2}=-\frac{11}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$y=7 y=-4$$