Help with solving this system of equations

y = (x + 3)^2 -4
y= 2x + 5

Choices: a (0,4) b (-4,0) c ( -4, -3) and (0,5) d (-3, -4) and (5, 0 )
Please help me with this. I have a final tomorrow and my answer is not working out.

Set the value of y in the second given equation to y in the first given equation:
2x + 5 = (x + 3)^2 -4
Expand on the right:
2x + 5 = x^2 + 6x + 9 - 4
Combine like terms:
x^2 + 4x = 0
Factor out the common factor:
x (x + 4) = 0
Use the zero product principle twice:
x = 0, -4
Substitute these values for x into y= 2x + 5 and evaluate for y twice:
y = 5, -3
So the two solutions are (0, 5) and (-4, -3)
Answer c.

What you want to do is make one of the variables cancel by manipulating one of the equations.
If it were me I would make the y cancel by multiplying the bottom equation by -1.
Now, y=(x+3)^2-4 and -y=-2x-5
Then you add the common numbers together.
So you get 0=(x+3)^2-2x-9
Then you use the quadratic formula or a calculator if so inclined and you find your x value.
When you do this you will get x=0.
Then plug 0 in one of the original equations, y=2(0)+5
So y=5.
Your answer is then c. Since (0,5) is one of the two points.