x^2+20x-0.125=0

Solution for x^2+20x-0.125=0 equation:

x^2+20x-0.125=0

a = 1; b = 20; c = -0.125;
Δ = b2-4ac
Δ = 202-4·1·(-0.125)
Δ = 400.5
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{-(20)-\sqrt{400.5}}{2*1}=\frac{-20-\sqrt{400.5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{400.5}}{2*1}=\frac{-20+\sqrt{400.5}}{2}
$