What is the solution of the linear-quadratic system of equations?

y=x^2+5x-3

y-x=2

-x=-y+2

x=y-2

now substitute iin the first one

y = (y-2) ^2+5 (y-2) - 3

y=y^2-4y+4 + 5y-10-3

y=y^2+y-9

y^2-9=0.

y^2=9

y=3

3-x=2

-x=2-3

-x=-1

x=1

the solution are x=1 and y=3

3=1^2+5 (1) - 3

3=1+5-3

3=6-3

3=3