Explanation:
To solve this equation, we need to consider two cases, one where x is positive and another where x is negative.

Case 1: x is positive
When x is positive, the equation becomes:
x(6x2 - 1) = 5x2
Simplifying this equation, we get:
6x3 - x = 5x2
Rearranging the terms, we get:
6x3 - 5x2 - x = 0
Factoring out x, we get:
x(6x2 - 5x - 1) = 0
Using the quadratic formula, we can solve for the roots of 6x2 - 5x - 1 = 0:
x = (5 ± √37) / 12
Since x is positive, the only valid solution is:
x = (5 + √37) / 12

Case 2: x is negative
When x is negative, the equation becomes:
-x(6x2 - 1) = 5x2
Simplifying this equation, we get:
-6x3 + x = 5x2
Rearranging the terms, we get:
6x3 + 5x2 - x = 0
Factoring out -x, we get:
-x(6x2 + 5x - 1) = 0
Using the quadratic formula, we can solve for the roots of 6x2 + 5x - 1 = 0:
x = (-5 ± √61) / 12
Since x is negative, the only valid solution is:
x = (-5 - √61) / 12

Therefore, the total number of solutions is 2:

(5 + √37) / 12 and (-5 - √61) / 12

However, we need to check if these solutions satisfy the original equation. We can do this by plugging in each solution for x and verifying that the equation holds true. When we do this, we find that the solution (-5 - √61) / 12 does not satisfy the original equation. Therefore, the correct answer is 1 solution:

(5 + √37) / 12.

Final answer: 1 solution.