Solve the inequality $x^2 - 3 > 0$

HINT

Remember $ab > 0$ if $a,b > 0$ or $a,b < 0$

You have $3$ intervals to consider: $(-\infty,-\sqrt3), (-\sqrt3,\sqrt3)$ and $(\sqrt3,\infty)$. The sign of the product $(x-\sqrt3)(x+\sqrt3)$ is constant on each interval. (That's by continuity, since the function $f(x)=x^2-3$ must pass through zero to change sign.)

Use a test point in each interval: \begin{align} (-\infty,-\sqrt3): f(x)&\gt0\\ (-\sqrt3,\sqrt3): f(x)&\lt0\\ (\sqrt3,\infty): f(x)&\gt0\end{align}

So actually you get $x\lt -\sqrt3 \color{blue}{\text{ or }} x\gt\sqrt3 $.