y > x^2
y < -2(x + 2)^2 + 6
Instead of inequalities, let's just set these equal and see if we can find the correct solutions
x^2 = -2(x + 2)^2 + 6
x^2 = -2(x^2 + 4x + 4) + 6
x^2 = -2x^2 -8x - 8 + 6
x^2 = - 2x^2 - 8x - 2 rearrange as
3x^2 + 8x + 2 = 0
Solving this with the Quadratic Formula, we get the solutions :
x = ( -4 - √10 ) / 3 ≈ -2.387
x = (-4 + √10 ) / 3 ≈ -0.279
We have three possible intervals for solutions
(-infinity, ≈ -2.87 ) U ( ≈-2.87, ≈ -0.279) U (≈ -0.279, infinity)
The way these usually work is that either the two outside intervals work or the middle interval does
Here's a graph to prove that the middle interval is correct :
https://www.desmos.com/calculator/uvqnynvqym
The solution area is where both graphs overlap