#### Answer

The solution set is $\left\{-\dfrac{8}{5}\right\}$.

#### Work Step by Step

Square both sides to obtain:
$x^2-x-4=(x+2)^2$
Use the rule $(a+b)^2 = a^2+2ab+b^2$ to obtain:
$x^2-x-4=x^2+2(x)(2) + 2^2
\\x^2-x-4=x^2+4x+4$
Put all terms on the left side of the equation. and then combine like terms.
Note that when a term is transferred to the other side of an equation, its sign/operation changes to its opposite.
$x^2-x-4-x^2-4x-4=0
\\(x^2-x^2)+(-x-4x) + (-4-4) = 0
\\-5x+(-8)=0
\\-5x-8=0$
Add $5x$ on both sides of the equation to obtain:
$-8 = 5x$
Divide 5 on both sides of the equation to obtain:
$\dfrac{-8}{5} = x$
Thus, the solution set is $\left\{-\dfrac{8}{5}\right\}$.