X^2+40x-329=0

Solution for X^2+40x-329=0 equation:

X^2+40X-329=0

a = 1; b = 40; c = -329;
Δ = b2-4ac
Δ = 402-4·1·(-329)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-54}{2*1}=\frac{-94}{2}
=-47
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+54}{2*1}=\frac{14}{2}
=7
$