$x<-4 \lor x > 1$

We have the inequality $$x^2-4 > -3 x$$ but first we solve the following equation: $x^2-4 = -3 x$, that is $x^2+3 x-4 =0$. We do this via the quadratic formula: $$\begin{aligned} x&=\frac{-3\pm \sqrt{(-3)^2-4 \cdot 1 \cdot -4}}{2}\\ \\ &=\frac{-3\pm \sqrt{25}}{2}\\ \\ &=\frac{-3\pm 5}{2}\end{aligned}$$ So $$x=-4\quad \text{or}\quad x=1$$ Now we explore where the inequality is true.
First we take a value $x<-4$, say $x=-6$. The value of the left-hand side of the inequality is then $$(-6)^2-4=32$$ The value of the right-hand side is $$-3 \cdot -6=18$$ So we have found for $x<-4$ that $x^2-4 > -3 x$.
Next we choose a value $-4<x<1$, say $x=-3$. The value of the left-hand side of the inequality is then $$(-3)^2-4=5$$ The value of the right-handside is $$-3\cdot -3=9$$ So we have found for $-4<x<1$ that $x^2-4 < -3 x$.
Finally we choose a value $x>1$, say $x=2$. The value of the left-hand side of the inequality is then $$(2)^2-4=0$$ The value of the right-hand side is $$-3 \cdot 2=-6$$ So we have found for $x>1$ that $x^2-4>-3 x$.

So we can conclude that $$x^2-4 > -3 x$$ when $x<-4$ or $x>1$.