-m^2-m+20=0

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

-m^2-m+20=0

We add all the numbers together, and all the variables

-1m^2-1m+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:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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