x^2+13+40x=0

Solution for x^2+13+40x=0 equation:

x^2+13+40x=0

a = 1; b = 40; c = +13;
Δ = b2-4ac
Δ = 402-4·1·13
Δ = 1548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1548}=\sqrt{36*43}=\sqrt{36}*\sqrt{43}=6\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-6\sqrt{43}}{2*1}=\frac{-40-6\sqrt{43}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+6\sqrt{43}}{2*1}=\frac{-40+6\sqrt{43}}{2}
$