Solve (x+2)^2+4=(x-1)^2+7x

$$( x + 2 ) ^ { 2 } + 4 = ( x - 1 ) ^ { 2 } + 7 x$$

Solve for x

$$x=7$$

Steps for Solving Linear Equation

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

$$x^{2}+4x+4+4=\left(x-1\right)^{2}+7x$$

Add 4 and 4 to get 8.

$$x^{2}+4x+8=\left(x-1\right)^{2}+7x$$

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

$$x^{2}+4x+8=x^{2}-2x+1+7x$$

Combine -2x and 7x to get 5x.

$$x^{2}+4x+8=x^{2}+5x+1$$

Subtract x^{2} from both sides.

$$x^{2}+4x+8-x^{2}=5x+1$$

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

$$4x+8=5x+1$$

Subtract 5x from both sides.

$$4x+8-5x=1$$

Combine 4x and -5x to get -x.

$$-x+8=1$$

Subtract 8 from both sides.

$$-x=1-8$$

Subtract 8 from 1 to get -7.

$$-x=-7$$

Multiply both sides by -1.

$$x=7$$