0=3x^2+20x-40

Solution for 0=3x^2+20x-40 equation:

0=3x^2+20x-40

We move all terms to the left:

0-(3x^2+20x-40)=0

We add all the numbers together, and all the variables

-(3x^2+20x-40)=0

We get rid of parentheses

-3x^2-20x+40=0

a = -3; b = -20; c = +40;
Δ = b2-4ac
Δ = -202-4·(-3)·40
Δ = 880
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:

$\sqrt{\Delta}=\sqrt{880}=\sqrt{16*55}=\sqrt{16}*\sqrt{55}=4\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{55}}{2*-3}=\frac{20-4\sqrt{55}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{55}}{2*-3}=\frac{20+4\sqrt{55}}{-6}
$