3x^2+6-99=0

Solution for 3x^2+6-99=0 equation:

3x^2+6-99=0

We add all the numbers together, and all the variables

3x^2-93=0

a = 3; b = 0; c = -93;
Δ = b2-4ac
Δ = 02-4·3·(-93)
Δ = 1116
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:

$\sqrt{\Delta}=\sqrt{1116}=\sqrt{36*31}=\sqrt{36}*\sqrt{31}=6\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{31}}{2*3}=\frac{0-6\sqrt{31}}{6}
=-\frac{6\sqrt{31}}{6}
=-\sqrt{31}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{31}}{2*3}=\frac{0+6\sqrt{31}}{6}
=\frac{6\sqrt{31}}{6}
=\sqrt{31}
$