n^2-41n+40=0

Solution for n^2-41n+40=0 equation:

n^2-41n+40=0

a = 1; b = -41; c = +40;
Δ = b2-4ac
Δ = -412-4·1·40
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-39}{2*1}=\frac{2}{2}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+39}{2*1}=\frac{80}{2}
=40
$