X^2+40x=-14

Solution for X^2+40x=-14 equation:

X^2+40X=-14

We move all terms to the left:

X^2+40X-(-14)=0

We add all the numbers together, and all the variables

X^2+40X+14=0

a = 1; b = 40; c = +14;
Δ = b2-4ac
Δ = 402-4·1·14
Δ = 1544
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:

$\sqrt{\Delta}=\sqrt{1544}=\sqrt{4*386}=\sqrt{4}*\sqrt{386}=2\sqrt{386}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{386}}{2*1}=\frac{-40-2\sqrt{386}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{386}}{2*1}=\frac{-40+2\sqrt{386}}{2}
$