Answer
$\bigg\{\frac{-1-\sqrt{17}}{2},3\bigg\}$
Work Step by Step
Taking the positive:
$$x+1=x^2-5$$
$$0=x^2-x-6$$
$$(x-3)(x+2)=0$$
$$x-3=0$$
$$x=3$$
$$x+2=0$$
$$x=-2$$
Checking:
For $x=3$:
$$|3+1|=3^2-5$$
$$4=4~True$$
Thus, $x=3$ is a solution.
For $x=-2$:
$$|-2+1|=(-2)^2-5$$
$$1=-1~False$$
Thus, $x=-2$ is not a solution.
Taking the negative:
$$x+1=-(x^2-5)$$
$$x+1=-x^2+5$$
$$x^2+x-4=0$$
Using the quadratic formula:
$$x=\frac{-1\pm\sqrt{(-1)^2-4(1)(-4)}}{2(1)}=\frac{-1\pm\sqrt{17}}{2}$$
$$x_1=\frac{-1+\sqrt{17}}{2}$$
$$x_2=\frac{-1-\sqrt{17}}{2}$$
Checking:
For $x=\frac{-1+\sqrt{17}}{2}$:
$$|\frac{-1+\sqrt{17}}{2}+1|=\left(\frac{-1+\sqrt{17}}{2}\right)^2-5$$
$$\frac{-1+\sqrt{17}}{2}+1=\frac{9-\sqrt{17}}{2}-5~False$$
Thus, $x=\frac{-1+\sqrt{17}}{2}$ is not a solution.
For $x=\frac{-1-\sqrt{17}}{2}$:
$$|\frac{-1-\sqrt{17}}{2}+1|=\left(\frac{-1-\sqrt{17}}{2}\right)^2-5$$
$$-\frac{-1-\sqrt{17}}{2}-1=\frac{9-\sqrt{17}}{2}-5~True$$
Thus, $x=\frac{-1-\sqrt{17}}{2}$ is a solution.
Therefore, the solution set is:
$$\bigg\{\frac{-1-\sqrt{17}}{2},3\bigg\}$$
