3x^2+20x-20=0

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

3x^2+20x-20=0

a = 3; b = 20; c = -20;
Δ = b2-4ac
Δ = 202-4·3·(-20)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{10}}{2*3}=\frac{-20-8\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{10}}{2*3}=\frac{-20+8\sqrt{10}}{6}
$