# How do you solve x^2+2x-3>0?

$\to \left(x - 1\right) \left(x + 3\right) = 0 \to x = 1 \mathmr{and} x = - 3$
For any $x$-value in between we get a negative $y$, so the answer must be:
$x < - 3 \mathmr{and} x > 1$