-n2+41n-400=0

Solution for -n2+41n-400=0 equation:

-n2+41n-400=0

We add all the numbers together, and all the variables

-1n^2+41n-400=0

a = -1; b = 41; c = -400;
Δ = b2-4ac
Δ = 412-4·(-1)·(-400)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-9}{2*-1}=\frac{-50}{-2}
=+25
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+9}{2*-1}=\frac{-32}{-2}
=+16
$