Solve (y-2)^2+48=(2-3y)^2

$$( y - 2 ) ^ { 2 } + 48 = ( 2 - 3 y ) ^ { 2 }$$

Solve for y

$$y=-2$$

$$y=3$$

Steps Using Factoring By Grouping

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

$$y^{2}-4y+4+48=\left(2-3y\right)^{2}$$

Add 4 and 48 to get 52.

$$y^{2}-4y+52=\left(2-3y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-3y\right)^{2}.

$$y^{2}-4y+52=4-12y+9y^{2}$$

Subtract 4 from both sides.

$$y^{2}-4y+52-4=-12y+9y^{2}$$

Subtract 4 from 52 to get 48.

$$y^{2}-4y+48=-12y+9y^{2}$$

Add 12y to both sides.

$$y^{2}-4y+48+12y=9y^{2}$$

Combine -4y and 12y to get 8y.

$$y^{2}+8y+48=9y^{2}$$

Subtract 9y^{2} from both sides.

$$y^{2}+8y+48-9y^{2}=0$$

Combine y^{2} and -9y^{2} to get -8y^{2}.

$$-8y^{2}+8y+48=0$$

Divide both sides by 8.

$$-y^{2}+y+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=1 ab=-6=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=3 b=-2$$

Rewrite -y^{2}+y+6 as \left(-y^{2}+3y\right)+\left(-2y+6\right).

$$\left(-y^{2}+3y\right)+\left(-2y+6\right)$$

Factor out -y in the first and -2 in the second group.

$$-y\left(y-3\right)-2\left(y-3\right)$$

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(-y-2\right)$$

To find equation solutions, solve y-3=0 and -y-2=0.

$$y=3 y=-2$$

Steps Using the Quadratic Formula

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

$$y^{2}-4y+4+48=\left(2-3y\right)^{2}$$

Add 4 and 48 to get 52.

$$y^{2}-4y+52=\left(2-3y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-3y\right)^{2}.

$$y^{2}-4y+52=4-12y+9y^{2}$$

Subtract 4 from both sides.

$$y^{2}-4y+52-4=-12y+9y^{2}$$

Subtract 4 from 52 to get 48.

$$y^{2}-4y+48=-12y+9y^{2}$$

Add 12y to both sides.

$$y^{2}-4y+48+12y=9y^{2}$$

Combine -4y and 12y to get 8y.

$$y^{2}+8y+48=9y^{2}$$

Subtract 9y^{2} from both sides.

$$y^{2}+8y+48-9y^{2}=0$$

Combine y^{2} and -9y^{2} to get -8y^{2}.

$$-8y^{2}+8y+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 8 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-8±\sqrt{8^{2}-4\left(-8\right)\times 48}}{2\left(-8\right)}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\left(-8\right)\times 48}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$y=\frac{-8±\sqrt{64+32\times 48}}{2\left(-8\right)}$$

Multiply 32 times 48.

$$y=\frac{-8±\sqrt{64+1536}}{2\left(-8\right)}$$

Add 64 to 1536.

$$y=\frac{-8±\sqrt{1600}}{2\left(-8\right)}$$

Take the square root of 1600.

$$y=\frac{-8±40}{2\left(-8\right)}$$

Multiply 2 times -8.

$$y=\frac{-8±40}{-16}$$

Now solve the equation y=\frac{-8±40}{-16} when ± is plus. Add -8 to 40.

$$y=\frac{32}{-16}$$

Divide 32 by -16.

$$y=-2$$

Now solve the equation y=\frac{-8±40}{-16} when ± is minus. Subtract 40 from -8.

$$y=-\frac{48}{-16}$$

Divide -48 by -16.

$$y=3$$

The equation is now solved.

$$y=-2 y=3$$

Steps for Completing the Square

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

$$y^{2}-4y+4+48=\left(2-3y\right)^{2}$$

Add 4 and 48 to get 52.

$$y^{2}-4y+52=\left(2-3y\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-3y\right)^{2}.

$$y^{2}-4y+52=4-12y+9y^{2}$$

Add 12y to both sides.

$$y^{2}-4y+52+12y=4+9y^{2}$$

Combine -4y and 12y to get 8y.

$$y^{2}+8y+52=4+9y^{2}$$

Subtract 9y^{2} from both sides.

$$y^{2}+8y+52-9y^{2}=4$$

Combine y^{2} and -9y^{2} to get -8y^{2}.

$$-8y^{2}+8y+52=4$$

Subtract 52 from both sides.

$$-8y^{2}+8y=4-52$$

Subtract 52 from 4 to get -48.

$$-8y^{2}+8y=-48$$

Divide both sides by -8.

$$\frac{-8y^{2}+8y}{-8}=-\frac{48}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$y^{2}+\frac{8}{-8}y=-\frac{48}{-8}$$

Divide 8 by -8.

$$y^{2}-y=-\frac{48}{-8}$$

Divide -48 by -8.

$$y^{2}-y=6$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-y+\left(-\frac{1}{2}\right)^{2}=6+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-y+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$y^{2}-y+\frac{1}{4}=\frac{25}{4}$$

Factor y^{2}-y+\frac{1}{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}.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$y-\frac{1}{2}=\frac{5}{2} y-\frac{1}{2}=-\frac{5}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$y=3 y=-2$$