Question 955009
{{{5x+(x^2-4)=20}}}
{{{x^2+5x-4=20}}}
{{{x^2+5x-24=0}}}
{{{(x+8)(x-3)=0}}}
Two solutions:
{{{x+8=0}}}
{{{x=-8}}}
Then,
{{{y=(-8)^2-4}}}
{{{y=64-4}}}
{{{y=60}}}
and
{{{x-3=0}}}
{{{x=3}}}
Then,
{{{y=(3)^2-4}}}
{{{y=9-4}}}
{{{y=5}}}
So the two solutions are (-8,60) and (3,5)