x2+40x-41=0

Solution for x2+40x-41=0 equation:

x2+40x-41=0

We add all the numbers together, and all the variables

x^2+40x-41=0

a = 1; b = 40; c = -41;
Δ = b2-4ac
Δ = 402-4·1·(-41)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-42}{2*1}=\frac{-82}{2}
=-41
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+42}{2*1}=\frac{2}{2}
=1
$