Expain to me how to solve

First factor out common x^2:

(3x^2 - 4x - 3)(x^2) = 0

Now realize that x=0 is a solution (x^2 is zero when x=0, and anything multiplied by 0 equals 0). So note this, then divide both sides of the equation by x^2 (note that by doing so we're implying that x is not 0 and therefore lose the solution x=0. If x=0 we cannot do the division. If this confuses you ask your teacher about it):

(3x^2 - 4x - 3) = 0

Use the quadratic equation to solve for x:

x = ( -(-4) +- sqrt((-4)^2 - 4(3)(-3)) ) / (2*3)
x = (4 +- sqrt(16+36)) / 6
x = (4 +- sqrt(52)) / 6
x = (4 + sqrt(52)) / 6 OR x = (4 - sqrt(52)) / 6

So we have 3 solutions and are done.

----EDIT----
The above answerer is technically correct since x=0 is actually a double zero. However usually double roots are only considered to be one solution unless you're wanting to be extremely precise with your language. If this confuses you you can ignore it for now.

The equation 3x^4 - 4x^3 - 3x^2 = 0 has four real solutions.

x^2*(3*x^2 - 4*x - 3) = 0

x1 = x2 = 0

3*x^2 - 4*x - 3 = 0

x3 = 2/3 + (1/3)*sqrt(13), x4 = 2/3 - (1/3)*sqrt(13)