-n^2+18n+40=0

Solution for -n^2+18n+40=0 equation:

-n^2+18n+40=0

We add all the numbers together, and all the variables

-1n^2+18n+40=0

a = -1; b = 18; c = +40;
Δ = b2-4ac
Δ = 182-4·(-1)·40
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*-1}=\frac{-40}{-2}
=+20
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*-1}=\frac{4}{-2}
=-2
$