3x^2+20x-11=0

Solution for 3x^2+20x-11=0 equation:

3x^2+20x-11=0

a = 3; b = 20; c = -11;
Δ = b2-4ac
Δ = 202-4·3·(-11)
Δ = 532
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:

$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{133}}{2*3}=\frac{-20-2\sqrt{133}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{133}}{2*3}=\frac{-20+2\sqrt{133}}{6}
$