0=3x^2+20x-100

Solution for 0=3x^2+20x-100 equation:

0=3x^2+20x-100

We move all terms to the left:

0-(3x^2+20x-100)=0

We add all the numbers together, and all the variables

-(3x^2+20x-100)=0

We get rid of parentheses

-3x^2-20x+100=0

a = -3; b = -20; c = +100;
Δ = b2-4ac
Δ = -202-4·(-3)·100
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-40}{2*-3}=\frac{-20}{-6}
=3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+40}{2*-3}=\frac{60}{-6}
=-10
$