Solve it like a normal quadratic equation and take the root of the roots
x^4 + 10x^2 + 25 = 0
Δ = b² - 4.a.c
Δ = 10² - 4 . 1 . 25
Δ = 100 - 4. 1 . 25
Δ = 0
1 real root.
In this case, x' = x'':
x = (-b +- √Δ)/2a
x' = (-10 + √0)/2.1
x'' = (-10 - √0)/2.1
x' = -10 / 2
x'' = -10 / 2
x' = -5
x'' = -5
Now take the root of this roots
√-5 = i√5
The root of this equation is x = ±i√5
