0=3x^2+20x-35

Solution for 0=3x^2+20x-35 equation:

0=3x^2+20x-35

We move all terms to the left:

0-(3x^2+20x-35)=0

We add all the numbers together, and all the variables

-(3x^2+20x-35)=0

We get rid of parentheses

-3x^2-20x+35=0

a = -3; b = -20; c = +35;
Δ = b2-4ac
Δ = -202-4·(-3)·35
Δ = 820
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:

$\sqrt{\Delta}=\sqrt{820}=\sqrt{4*205}=\sqrt{4}*\sqrt{205}=2\sqrt{205}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{205}}{2*-3}=\frac{20-2\sqrt{205}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{205}}{2*-3}=\frac{20+2\sqrt{205}}{-6}
$