Solve x^2+3x+8=(x+2)^2-19

$$x ^ { 2 } + 3 x + 8 = ( x + 2 ) ^ { 2 } - 19$$

Solve for x

$$x=23$$

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}.

$$x^{2}+3x+8=x^{2}+4x+4-19$$

Subtract 19 from 4 to get -15.

$$x^{2}+3x+8=x^{2}+4x-15$$

Subtract x^{2} from both sides.

$$x^{2}+3x+8-x^{2}=4x-15$$

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

$$3x+8=4x-15$$

Subtract 4x from both sides.

$$3x+8-4x=-15$$

Combine 3x and -4x to get -x.

$$-x+8=-15$$

Subtract 8 from both sides.

$$-x=-15-8$$

Subtract 8 from -15 to get -23.

$$-x=-23$$

Multiply both sides by -1.

$$x=23$$