-a^2+a+20=0

Solution for -a^2+a+20=0 equation:

-a^2+a+20=0

We add all the numbers together, and all the variables

-1a^2+a+20=0

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

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