Solve (y+4)^2=2y^2-3y+44

$$( y + 4 ) ^ { 2 } = 2 y ^ { 2 } - 3 y + 44$$

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}.

$$y^{2}+8y+16=2y^{2}-3y+44$$

Subtract 2y^{2} from both sides.

$$y^{2}+8y+16-2y^{2}=-3y+44$$

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

$$-y^{2}+8y+16=-3y+44$$

Add 3y to both sides.

$$-y^{2}+8y+16+3y=44$$

Combine 8y and 3y to get 11y.

$$-y^{2}+11y+16=44$$

Subtract 44 from both sides.

$$-y^{2}+11y+16-44=0$$

Subtract 44 from 16 to get -28.

$$-y^{2}+11y-28=0$$

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.

$$a+b=11 ab=-\left(-28\right)=28$$

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 28.

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

Calculate the sum for each pair.

$$1+28=29 2+14=16 4+7=11$$

The solution is the pair that gives sum 11.

$$a=7 b=4$$

Rewrite -y^{2}+11y-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}.

$$y^{2}+8y+16=2y^{2}-3y+44$$

Subtract 2y^{2} from both sides.

$$y^{2}+8y+16-2y^{2}=-3y+44$$

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

$$-y^{2}+8y+16=-3y+44$$

Add 3y to both sides.

$$-y^{2}+8y+16+3y=44$$

Combine 8y and 3y to get 11y.

$$-y^{2}+11y+16=44$$

Subtract 44 from both sides.

$$-y^{2}+11y+16-44=0$$

Subtract 44 from 16 to get -28.

$$-y^{2}+11y-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 11 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-11±\sqrt{11^{2}-4\left(-1\right)\left(-28\right)}}{2\left(-1\right)}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\left(-1\right)\left(-28\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-11±\sqrt{121+4\left(-28\right)}}{2\left(-1\right)}$$

Multiply 4 times -28.

$$y=\frac{-11±\sqrt{121-112}}{2\left(-1\right)}$$

Add 121 to -112.

$$y=\frac{-11±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$y=\frac{-11±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-11±3}{-2}$$

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

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

Divide -8 by -2.

$$y=4$$

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

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

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}.

$$y^{2}+8y+16=2y^{2}-3y+44$$

Subtract 2y^{2} from both sides.

$$y^{2}+8y+16-2y^{2}=-3y+44$$

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

$$-y^{2}+8y+16=-3y+44$$

Add 3y to both sides.

$$-y^{2}+8y+16+3y=44$$

Combine 8y and 3y to get 11y.

$$-y^{2}+11y+16=44$$

Subtract 16 from both sides.

$$-y^{2}+11y=44-16$$

Subtract 16 from 44 to get 28.

$$-y^{2}+11y=28$$

Divide both sides by -1.

$$\frac{-y^{2}+11y}{-1}=\frac{28}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{11}{-1}y=\frac{28}{-1}$$

Divide 11 by -1.

$$y^{2}-11y=\frac{28}{-1}$$

Divide 28 by -1.

$$y^{2}-11y=-28$$

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.

$$y^{2}-11y+\left(-\frac{11}{2}\right)^{2}=-28+\left(-\frac{11}{2}\right)^{2}$$

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

$$y^{2}-11y+\frac{121}{4}=-28+\frac{121}{4}$$

Add -28 to \frac{121}{4}.

$$y^{2}-11y+\frac{121}{4}=\frac{9}{4}$$

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}.

$$\left(y-\frac{11}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{11}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{11}{2} to both sides of the equation.

$$y=7 y=4$$