20x^2+0.8x-0.8=0

Solution for 20x^2+0.8x-0.8=0 equation:

20x^2+0.8x-0.8=0

a = 20; b = 0.8; c = -0.8;
Δ = b2-4ac
Δ = 0.82-4·20·(-0.8)
Δ = 64.64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.8)-\sqrt{64.64}}{2*20}=\frac{-0.8-\sqrt{64.64}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.8)+\sqrt{64.64}}{2*20}=\frac{-0.8+\sqrt{64.64}}{40}
$