93=3u^2+12

Solution for 93=3u^2+12 equation:

93=3u^2+12

We move all terms to the left:

93-(3u^2+12)=0

We get rid of parentheses

-3u^2-12+93=0

We add all the numbers together, and all the variables

-3u^2+81=0

a = -3; b = 0; c = +81;
Δ = b2-4ac
Δ = 02-4·(-3)·81
Δ = 972
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{3}}{2*-3}=\frac{0-18\sqrt{3}}{-6}
=-\frac{18\sqrt{3}}{-6}
=-\frac{3\sqrt{3}}{-1}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{3}}{2*-3}=\frac{0+18\sqrt{3}}{-6}
=\frac{18\sqrt{3}}{-6}
=\frac{3\sqrt{3}}{-1}
$