Solve system by elimination y=x^2 y=x+2 show your work

We have that
y=x²> equation 1
y=x+2> equation 2

multiply equation 1 by -1
-y=-x²

add equation 1 and equation 2
-y=-x²
 y=x+2

0=-x²+x+2> -x²+x+2=0> x²-x-2=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(x²-x)=2
Complete the square. Remember to balance the equation by adding the same constants to each side
(x²-x+0.5²)=2+0.5²

Rewrite as perfect squares

(x-0.5)²=2+0.5²
(x-0.5)²=2.25> (x-0.5)=(+/-)√2.25> (x-0.5)=(+/-)1.5
x1=1.5+0.5> x1=2
x2=-1.5+0.5 > x2=-1

for x=2
y=x²> y=2²> y=4
the point is (2,4)

for x=-1
y=x²> y=(-1)²---> y=1
the point is (-1,1)

the answer is
the solution of the system are the points
(2,4) and (-1,1)