Solve (-3-y)^2-(2-y)^2=(y+4)*(4-y)+y(y+1)-2

$$( - 3 - y ) ^ { 2 } - ( 2 - y ) ^ { 2 } = ( y + 4 ) \cdot ( 4 - y ) + y ( y + 1 ) - 2$$

Solve for y

$$y=1$$

Steps for Solving Linear Equation

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

$$9+6y+y^{2}-\left(2-y\right)^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

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

$$9+6y+y^{2}-\left(4-4y+y^{2}\right)=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

To find the opposite of 4-4y+y^{2}, find the opposite of each term.

$$9+6y+y^{2}-4+4y-y^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Subtract 4 from 9 to get 5.

$$5+6y+y^{2}+4y-y^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Combine 6y and 4y to get 10y.

$$5+10y+y^{2}-y^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

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

$$5+10y=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Consider \left(y+4\right)\left(4-y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.

$$5+10y=16-y^{2}+y\left(y+1\right)-2$$

Use the distributive property to multiply y by y+1.

$$5+10y=16-y^{2}+y^{2}+y-2$$

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

$$5+10y=16+y-2$$

Subtract 2 from 16 to get 14.

$$5+10y=14+y$$

Subtract y from both sides.

$$5+10y-y=14$$

Combine 10y and -y to get 9y.

$$5+9y=14$$

Subtract 5 from both sides.

$$9y=14-5$$

Subtract 5 from 14 to get 9.

$$9y=9$$

Divide both sides by 9.

$$y=\frac{9}{9}$$

Divide 9 by 9 to get 1.

$$y=1$$