Originally Posted by

**LoveBeachForever** Okay so if $\displaystyle 0=y^2-y-7$

then I can use the quadratic and get $\displaystyle y=\frac{1 \pm \sqrt 29}{2}$

which is $\displaystyle y=3.193$ or $\displaystyle -2.193$

using that and plugging it back in i get:

$\displaystyle 3.193=2-x^2$

$\displaystyle x^2=1.193$

$\displaystyle x=1.092$ so the solution is at (3.193,1.092) correct?

then for the other one,

$\displaystyle -2.193=2-x^2$

but $\displaystyle x^2=-4.193$ which leaves me with an imaginary number. So I'm assuming that it means that just the other one is the solution. Am I right?