Solve (3y+3)^2=5y^2+(2y-3)^2

$$( 3 y + 3 ) ^ { 2 } = 5 y ^ { 2 } + ( 2 y - 3 ) ^ { 2 }$$

Solve for y

$$y=0$$

Steps for Solving Linear Equation

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

$$9y^{2}+18y+9=5y^{2}+\left(2y-3\right)^{2}$$

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

$$9y^{2}+18y+9=5y^{2}+4y^{2}-12y+9$$

Combine 5y^{2} and 4y^{2} to get 9y^{2}.

$$9y^{2}+18y+9=9y^{2}-12y+9$$

Subtract 9y^{2} from both sides.

$$9y^{2}+18y+9-9y^{2}=-12y+9$$

Combine 9y^{2} and -9y^{2} to get 0.

$$18y+9=-12y+9$$

Add 12y to both sides.

$$18y+9+12y=9$$

Combine 18y and 12y to get 30y.

$$30y+9=9$$

Subtract 9 from both sides.

$$30y=9-9$$

Subtract 9 from 9 to get 0.

$$30y=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 30 is not equal to 0, y must be equal to 0.

$$y=0$$