The equation to be solved is: `5(x^2 + 1/x^2) - 3(x^4 + 1/x^4) = 0`

Use the relation: `x^4 + 1/x^4 = (x^2 + 1/x^2)^2 - 2`

=> `5(x^2 + 1/x^2) - 3((x^2 + 1/x^2)^2 - 2) = 0`

Let `y = x^2 + 1/x^2`

=> 5y - 3(y^2 - 2) = 0

=> 3y^2 - 5y - 6 = 0

=> y1 = `(5 - sqrt 97)/6`

y2 = `(5 + sqrt 97)/6`

This gives the equations: x^2 + 1/x^2 = `(5 - sqrt 97)/6` and x^2 + 1/x^2 = `(5 +sqrt 97)/6`

Solving these equations gives the value of x.

This can be done by substituting x^2 = y and solving the equation y +1/y = `(5- sqrt 97)/6` and y + 1/y = `(5 + sqrt 97)/6`

Again for each value of y there will be 2 values of x.

**The total number of solutions of the equation are 8 and they are:**

x1 = -sqrt(4*sqrt(5*sqrt(97)-11)+2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))

x2 = sqrt(4*sqrt(5*sqrt(97)-11)+2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))

x3 = -sqrt(-4*sqrt(5*sqrt(97)-11)+2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))

x4 = sqrt(-4*sqrt(5*sqrt(97)-11)+2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))

x5 = -sqrt(4*sqrt(5*sqrt(97)+11)*i-2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))

x6 = sqrt(4*sqrt(5*sqrt(97)+11)*i-2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))

x7 = -sqrt(-4*sqrt(5*sqrt(97)+11)*i-2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))

x8 = sqrt(-4*sqrt(5*sqrt(97)+11)*i-2^(3/2)*sqrt(97)+5*2^(3/2))/(2^(7/4)*sqrt(3))