Question 433391
Using the substitution method we can write:

{{{-x+2=x^2+3x+6}}}, we solve this quadratic equation

{{{x^2+4x+4=0}}}, as you see the left side is the square of sum.

{{{(x+2)^2=0}}}, x+2=0 => x=-2. Substitute this value of x on the second equation 

of the system to find y.

{{{y=-(-2)+2=4}}},thus the solution is the point (-2, 4).