Solve the equation x^4 - 5x^2 - 36 = 0

We'll substitute x^2 = t and we'll re-write the equation
in t;

t^2 - 5t - 36 = 0

We'll
solve the quadratic, using the formula:

t1 = [-b+sqrt(b^2 -
4ac)]/2a

t2 = [-b-sqrt(b^2 -
4ac)]/2a

We'll identify the coefficients
a,b,c:

a = 1 , b = -5 , c =
-36

t1 = [5+sqrt(25 +
144)]/2

t1 = (5+13)/2

t1 =
9

t2 = (5-13)/2

t2 =
-4

But x^2 = t1 => x^2 = 9 => x1 = 3 and x2 =
-3

x^2 = t2 => x^2 = -4 => x3 = 2i and x4 =
-2i

The solutions of the equation are: {-3 ;
3 ; -2i ; 2i}.