#### Answer

The solution set of this problem, written in standard form, is $$\Big\{-\frac{1}{2}\pm\frac{\sqrt3}{2}i\Big\}$$

#### Work Step by Step

$$x^2+1=-x$$
The equation is not in standard form, so first, we need to bring it back to standard form:
$$x^2+x+1=0$$
Now the equation is in standard form, the quadratic formula can be applied.
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
As $a=1, b=1, c=1$
$$x=\frac{-1\pm\sqrt{1^2-4\times1\times1}}{2\times1}$$
$$x=\frac{-1\pm\sqrt{1-4}}{2}$$
$$x=\frac{-1\pm\sqrt{-3}}{2}$$
Now we rewrite $\sqrt{-3}=i\sqrt{3}$
$$x=\frac{-1\pm i\sqrt3}{2}$$
The solution set of this problem, written in standard form, is $$\Big\{-\frac{1}{2}\pm\frac{\sqrt3}{2}i\Big\}$$