20x^2+6x-20=0

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

20x^2+6x-20=0

a = 20; b = 6; c = -20;
Δ = b2-4ac
Δ = 62-4·20·(-20)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{409}}{2*20}=\frac{-6-2\sqrt{409}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{409}}{2*20}=\frac{-6+2\sqrt{409}}{40}
$