Question 59973
Solve the system of equations by substitution

y = x^2 + 2x - 2   and  y - 3x + 4 <---you have a type-o here, I'm assuming that the - is supposed to be =.
{{{y=x^2+2x-2}}} and {{{y=3x+4}}}
Se the two y's equal to each other:
{{{x^2+2x-2=3x+4}}}  Set = to 0
{{{x^2+2x-3x-2-4=3x-3x+4-4}}}
{{{x^2-x-6=0}}}  Factor
(x-3)(x+2)=0  Set each parenthesis = to 0.
x-3=0 and x+2=0
x-3+3=0+3 and x+2-2=0-2
x=3 and x=-2
Substitute 3 in for x in the linear equation because it's easiest.
y=3(3)+4
y=9+4
y=13  One solution is (3,13)
Sustitute -2 in for x in the same equation:
y=3(-2)+4
y=-6+4
y=-2  The second solution is (-2,-2)
Graphically, you can see the two points of intersection:
{{{graph(300,200,-10,10,-10,15,x^2+2x-2,3x+4)}}}
Happy Calculating!!!