100=3x^2+5

Solution for 100=3x^2+5 equation:

100=3x^2+5

We move all terms to the left:

100-(3x^2+5)=0

We get rid of parentheses

-3x^2-5+100=0

We add all the numbers together, and all the variables

-3x^2+95=0

a = -3; b = 0; c = +95;
Δ = b2-4ac
Δ = 02-4·(-3)·95
Δ = 1140
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:

$\sqrt{\Delta}=\sqrt{1140}=\sqrt{4*285}=\sqrt{4}*\sqrt{285}=2\sqrt{285}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{285}}{2*-3}=\frac{0-2\sqrt{285}}{-6}
=-\frac{2\sqrt{285}}{-6}
=-\frac{\sqrt{285}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{285}}{2*-3}=\frac{0+2\sqrt{285}}{-6}
=\frac{2\sqrt{285}}{-6}
=\frac{\sqrt{285}}{-3}
$