Solve 36(y+3)=(14-2y)^2

$$36 ( y + 3 ) = ( 14 - 2 y ) ^ { 2 }$$

Solve for y

$$y=1$$

$$y=22$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 36 by y+3.

$$36y+108=\left(14-2y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-2y\right)^{2}.

$$36y+108=196-56y+4y^{2}$$

Subtract 196 from both sides.

$$36y+108-196=-56y+4y^{2}$$

Subtract 196 from 108 to get -88.

$$36y-88=-56y+4y^{2}$$

Add 56y to both sides.

$$36y-88+56y=4y^{2}$$

Combine 36y and 56y to get 92y.

$$92y-88=4y^{2}$$

Subtract 4y^{2} from both sides.

$$92y-88-4y^{2}=0$$

Divide both sides by 4.

$$23y-22-y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-y^{2}+23y-22=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-22. To find a and b, set up a system to be solved.

$$a+b=23 ab=-\left(-22\right)=22$$

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 22.

$$1,22 2,11$$

Calculate the sum for each pair.

$$1+22=23 2+11=13$$

The solution is the pair that gives sum 23.

$$a=22 b=1$$

Rewrite -y^{2}+23y-22 as \left(-y^{2}+22y\right)+\left(y-22\right).

$$\left(-y^{2}+22y\right)+\left(y-22\right)$$

Factor out -y in -y^{2}+22y.

$$-y\left(y-22\right)+y-22$$

Factor out common term y-22 by using distributive property.

$$\left(y-22\right)\left(-y+1\right)$$

To find equation solutions, solve y-22=0 and -y+1=0.

$$y=22 y=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 36 by y+3.

$$36y+108=\left(14-2y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-2y\right)^{2}.

$$36y+108=196-56y+4y^{2}$$

Subtract 196 from both sides.

$$36y+108-196=-56y+4y^{2}$$

Subtract 196 from 108 to get -88.

$$36y-88=-56y+4y^{2}$$

Add 56y to both sides.

$$36y-88+56y=4y^{2}$$

Combine 36y and 56y to get 92y.

$$92y-88=4y^{2}$$

Subtract 4y^{2} from both sides.

$$92y-88-4y^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4y^{2}+92y-88=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 92 for b, and -88 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-92±\sqrt{92^{2}-4\left(-4\right)\left(-88\right)}}{2\left(-4\right)}$$

Square 92.

$$y=\frac{-92±\sqrt{8464-4\left(-4\right)\left(-88\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$y=\frac{-92±\sqrt{8464+16\left(-88\right)}}{2\left(-4\right)}$$

Multiply 16 times -88.

$$y=\frac{-92±\sqrt{8464-1408}}{2\left(-4\right)}$$

Add 8464 to -1408.

$$y=\frac{-92±\sqrt{7056}}{2\left(-4\right)}$$

Take the square root of 7056.

$$y=\frac{-92±84}{2\left(-4\right)}$$

Multiply 2 times -4.

$$y=\frac{-92±84}{-8}$$

Now solve the equation y=\frac{-92±84}{-8} when ± is plus. Add -92 to 84.

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

Divide -8 by -8.

$$y=1$$

Now solve the equation y=\frac{-92±84}{-8} when ± is minus. Subtract 84 from -92.

$$y=-\frac{176}{-8}$$

Divide -176 by -8.

$$y=22$$

The equation is now solved.

$$y=1 y=22$$

Steps for Completing the Square

Use the distributive property to multiply 36 by y+3.

$$36y+108=\left(14-2y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-2y\right)^{2}.

$$36y+108=196-56y+4y^{2}$$

Add 56y to both sides.

$$36y+108+56y=196+4y^{2}$$

Combine 36y and 56y to get 92y.

$$92y+108=196+4y^{2}$$

Subtract 4y^{2} from both sides.

$$92y+108-4y^{2}=196$$

Subtract 108 from both sides.

$$92y-4y^{2}=196-108$$

Subtract 108 from 196 to get 88.

$$92y-4y^{2}=88$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4y^{2}+92y=88$$

Divide both sides by -4.

$$\frac{-4y^{2}+92y}{-4}=\frac{88}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$y^{2}+\frac{92}{-4}y=\frac{88}{-4}$$

Divide 92 by -4.

$$y^{2}-23y=\frac{88}{-4}$$

Divide 88 by -4.

$$y^{2}-23y=-22$$

Divide -23, the coefficient of the x term, by 2 to get -\frac{23}{2}. Then add the square of -\frac{23}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-23y+\left(-\frac{23}{2}\right)^{2}=-22+\left(-\frac{23}{2}\right)^{2}$$

Square -\frac{23}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-23y+\frac{529}{4}=-22+\frac{529}{4}$$

Add -22 to \frac{529}{4}.

$$y^{2}-23y+\frac{529}{4}=\frac{441}{4}$$

Factor y^{2}-23y+\frac{529}{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}.

$$\left(y-\frac{23}{2}\right)^{2}=\frac{441}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{23}{2}\right)^{2}}=\sqrt{\frac{441}{4}}$$

Simplify.

$$y-\frac{23}{2}=\frac{21}{2} y-\frac{23}{2}=-\frac{21}{2}$$

Add \frac{23}{2} to both sides of the equation.

$$y=22 y=1$$