Assuming that you mean
[tex]-12x^2+11x-3=0[/tex]
We can solve the equation with the usual quadratic formula:
[tex]ax^2+bx+c=0\iff x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In your case, the coefficients are
[tex]a=-12,\quad b=11,\quad c=-3[/tex]
So, the quadratic formula becomes
[tex]x=\dfrac{-11\pm\sqrt{121-144}}{24}[/tex]
This means that the equation has no real roots, because the quantity under square root is negative, and thus we can't evaluate the roots. If you're looking for complex roots, we have
[tex]\dfrac{-11\pm\sqrt{121-144}}{24}=\dfrac{-11\pm\sqrt{-23}}{24} \implies x_1 = \dfrac{-11+i\sqrt{23}}{24},\quad x_2 = \dfrac{-11-i\sqrt{23}}{24}[/tex]
