20x^2+10x-5=0

Solution for 20x^2+10x-5=0 equation:

20x^2+10x-5=0

a = 20; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·20·(-5)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{5}}{2*20}=\frac{-10-10\sqrt{5}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{5}}{2*20}=\frac{-10+10\sqrt{5}}{40}
$