x^2+20x-5^-3=0

Solution for x^2+20x-5^-3=0 equation:

x^2+20x-5^-3=0

We add all the numbers together, and all the variables

x^2+20x=0

a = 1; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·1·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*1}=\frac{-40}{2}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*1}=\frac{0}{2}
=0
$