Solve for real and complex solutions.Integral of 4x^3+6x =10

To determine all real and complex solutions of the
equation, we'll have to determine the indefinite integral from the left side, for the
beginning.

Int (4x^3+6x)dx = 4Int x^3 dx + 6Int x
dx

Int (4x^3+6x)dx = 4x^4/4 + 6x^2/2 +
C

Int (4x^3+6x)dx = x^4 +
3x^2

We'll substitute the integral from the left side by
it's result:

x^4 + 3x^2 =
10

We'll subtract 10 both
sides:

x^4 + 3x^2 - 10 = 0

It
is a bi-quadratic equation. We'll put x^2 = t.

t^2 + 3t -
10 = 0

We'll apply quadratic
formula:

t1 =
[-3+sqrt(9+40)]/2

t1 =
(-3+7)/2

t1 = 2

t2 =
(-3-7)/2

t2 = -5

x^2 = t1
=> x^2 = 2

We'll take square root both
sides:

x1 = -sqrt2

x2 =
sqrt2

x^2 = t2 => x^2 =
-5

We'll take square root both
sides:

x3 = -isqrt5

x4 =
isqrt5

The real and complex roots of the
given equation are; {-sqrt2 ; sqrt2 ; -isqrt5 ;
isqrt5}.