Help for two values of x

Question

I'm looking for help. Even if you just tell me the process rather than the answer.

Given that $y=10-3x^2$, find two values of $x$ for which $y=-17$.

How would I go about answering this?

Answer 1

Set $y = -17$, getting

$$-17 = 10 - 3x^2$$

Rearrange the equation to solve for $x^2$, and if possible take a square root.

Answer 2

$$-17 = y = 10 - 3x^2$$ Rearranging ($x$'es to one side, the rest to the other) gives $$3x^2 = 27$$ $$x^2 = 9$$ Taking the square root (which gives us two solutions) yields $$x = \pm \sqrt{9} = \pm 3$$