Question 379117
You're solving for the intersection of a line with a parabola.
.
.
.
1.{{{y+1=x^2+2x}}} 
2.{{{y+2x=4}}}
Subtract eq. 2 from eq. 1,
{{{y+1-y-2x=x^2+2x-4}}}
{{{1-2x=x^2+2x-4}}}
{{{x^2+4x-5=0}}}
{{{(x+5)(x-1)=0}}}
Two solutions:
{{{x+5=0}}}
{{{x=-5}}}
Then from eq. 2,
{{{y-10=4}}}
{{{y=14}}}
({{{-5}}},{{{14}}})
.
.
.
{{{x-1=0}}}
{{{x=1}}}
Then from eq. 2,
{{{y+2=4}}}
{{{y=2}}}
({{{1}}},{{{2}}})
.
.
.
{{{drawing(300,300,-9,9,-2,16,grid(1),circle(-5,14,0.3),circle(1,2,0.3),graph(300,300,-9,9,-2,16,0,x^2+2x-1,4-2x))}}}