93=3x^2+12

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



93=3x^2+12
We move all terms to the left:
93-(3x^2+12)=0
We get rid of parentheses
-3x^2-12+93=0
We add all the numbers together, and all the variables
-3x^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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$x_{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} $
$x_{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} $

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