Solve (x+3)(x-2)-(x+1)^2=1

$$( x + 3 ) ( x - 2 ) - ( x + 1 ) ^ { 2 } = 1$$

Solve for x

$$x=-8$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+3 by x-2 and combine like terms.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$x^{2}+x-6-x^{2}-2x-1=1$$

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

$$x-6-2x-1=1$$

Combine x and -2x to get -x.

$$-x-6-1=1$$

Subtract 1 from -6 to get -7.

$$-x-7=1$$

Add 7 to both sides.

$$-x=1+7$$

Add 1 and 7 to get 8.

$$-x=8$$

Multiply both sides by -1.

$$x=-8$$