Question

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

Answer 1

You first find the zero-points, and see what happens in-between

Explanation:

`->(x-1)(x+3)=0->x=1orx=-3`

For any `x`-value in between we get a negative `y`, so the answer must be:
`x<-3orx>1`
graph{x^2+2x-3 [-10, 10, -5, 5]}