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Use the substitution method to find the solutions to the system of equations. y=-x y=x^2-3
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-x = x^2 - 3 We can equal Y's.
y=-x y=x^2-3 y = y -x = x^2 - 3 0 = x^2 + x - 3 Use the quadratic formula.\[x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}\]\[x = \frac {-1 \pm \sqrt {1 + 12}}{2}\]\[x = \frac {1}{2}(-1 - \sqrt {13}), x = \frac {1}{2} (-1 + \sqrt {13})\]
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