Considering the equation x⁴=-7+24i, it is probably easier to first solve for x², and then for x. Let a+bi=x²

(x²)²=(a+bi)²=a²-b²+2abi

We know, then, that a²-b² is the real coefficient, -7.

Now, our original complex vector has a magnitude of √((a²-b²)²+(2ab)²)=a²+b², and also √((-7)²+24²)=25

So, a²+b²=25,
and a²-b²=-7

Adding the two equations yields

2a²=18
a²=9
a=3 or a=-3

Now, since 2ab=24, b=4 or b=-4

Basically, the two square roots of -7+24i are 3+4i and -(3+4i), so our fourth roots of -7+24i are going to be both square roots of 3+4i and i multiplied by both square roots of 3+4i.

Following a similar procedure, let the square root of 3+4i equal c+di.
We know that c²-d²=3, and that the magnitude of 3+4i, 5, equals c²+d²

c²-d²=3
c²+d²=5
2c²=8
c=2 or c=-2
2cd=4
d=1 or d=-1

So the square roots of 3+4i are 2+i and -2-i, and the square roots of -(3+4i) are -1+2i and 1-2i.

It's worth noting that, by the principle derived for finding square roots, it must be true that every time a complex number with integer coefficients is squared, a "pythagorean triple" is formed. For example,

(4+i)²=15+8i

8²+15²=17²

(8+3i)²=55+48i

48²+55²=73²