20x^2+1x-1=0

Solution for 20x^2+1x-1=0 equation:

20x^2+1x-1=0

We add all the numbers together, and all the variables

20x^2+x-1=0

a = 20; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·20·(-1)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*20}=\frac{-10}{40}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*20}=\frac{8}{40}
=1/5
$