Find the points of intersection, please help

x^2 + y^2 = 5
x - y = 1

1) x^2 + y^2 = 5
2) x - y = 1 or x = y + 1
substitute eq(2) into eq(1):
(y + 1)^2 + y^2 = 5
y^2 + 2y + 1 + y^2 = 5
2y^2 + 2y + 1 = 5
2y^2 + 2y - 4 = 0
using Quadratic Equation:
y = (-2 +/- SQR(4 - 4(2)(-4)))/2(2)
y = (-2 +/- 6)/4
y = -2, 1
using eq(2):
x = -2 + 1 = -1, x = 2
So the points of intersection are:
(-1,-2) and (2,1)

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