#### Answer

(0, 0) and (2, -4)

#### Work Step by Step

For this problem,
Equation 1: $x^{2}$ + y = 0
Equation 2: $x^{2}$ - 4x - y = 0
For substitution we need to solve for a variable. In this problem, we will solve for y in Equation 1 to get Equation 3:
y = -$x^{2}$ (Equation 3)
We then substitute that into Equation 2 and solve for x:
$x^{2}$ - 4x - (-$x^{2}$) = 0
$x^{2}$ - 4x + $x^{2}$ = 0
2$x^{2}$ - 4x = 0
2x(x -2) = 0
x = 0 and x = 2
We then use Equation 3:
y = -$0^{2}$ = 0
y = -$2^{2}$ = -4
Therefore (0, 0) and (2, -4) are solutions. Furthermore, the two graphs cross at (0, 0) and (2, -4), so the answer is confirmed.