>>1874740>>1874807If you square both sides, you get:
x^4 - 8x^2 + x + 12 = 0
You know you can break it down into two quadratics:
(a*x^2 + b*x + c)(p*x^2 + q*x + n) = 0
If you simplify the thing above, you get:
apx^4 +
aqx^3 +
bpx^3 +
anx^2 +
bqx^2 +
cpx^2 +
bnx +
cqx +
cn +
This gives you some useful constraints on the coefficients, mainly:
aq + bp = 0
ap = 1
bnx + cqx = 1
cn = 12
With a bit of trial and error you arrive at the following factoring:
(x^2 - x - 3)(x^2 + x - 4) = 0
Now you can solve each quadratic and check which solutions are valid for the original equation.