6x^2+20x-5=0

Solution for 6x^2+20x-5=0 equation:

6x^2+20x-5=0

a = 6; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·6·(-5)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{130}}{2*6}=\frac{-20-2\sqrt{130}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{130}}{2*6}=\frac{-20+2\sqrt{130}}{12}
$