Finding points of intersection of two curves.

Hey there, i've got two curves:
f(x) = (1/2)x^2 + 2
g(x) = 8 - x^4

I know I need to make g(x) = f(x), but its not working out. Just wondering if theres something i'm missing.

-Thanks

Setting f(x) = g(x), we have

(1/2)x^2 + 2 = 8 - x^4
x^2 + 4 = 16 - 2x^4
2x^4 + x^2 - 12 = 0

Let u = x^2

2u^2 + u - 12 = 0

Using the quadratic formula,

u = [-1 +/- sqrt(1^2 - 4(2)(-12))]/(2(2))
u = [-1 +/- sqrt(1 + 96)]/4
u = [-1 +/- sqrt(97)]/4

But since u = x^2, u must be nonnegative for there to be a real solution for x, so we must discard
u = [-1 - sqrt(97)]/4 < 0

We are left with
x^2 = u = [-1 + sqrt(97)]/4
x = +/-sqrt(-1 + sqrt(97))/2

Evaluating f(x) when x = +/-sqrt(-1 + sqrt(97))/2, we have

f(x) = (1/2)x^2 + 2
= (1/2)[+/-sqrt(-1 + sqrt(97))/2]^2 + 2
= (1/2)[-1 + sqrt(97)]/4 + 2
= [-1 + sqrt(97)]/8 + 2
= [15 + sqrt(97)]/8

Therefore the points of intersection are

(sqrt(-1 + sqrt(97))/2, [15 + sqrt(97)]/8) AND (-sqrt(-1 + sqrt(97))/2, [15 + sqrt(97)]/8)