X^2+40x-15=40

Solution for X^2+40x-15=40 equation:

X^2+40X-15=40

We move all terms to the left:

X^2+40X-15-(40)=0

We add all the numbers together, and all the variables

X^2+40X-55=0

a = 1; b = 40; c = -55;
Δ = b2-4ac
Δ = 402-4·1·(-55)
Δ = 1820
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:

$\sqrt{\Delta}=\sqrt{1820}=\sqrt{4*455}=\sqrt{4}*\sqrt{455}=2\sqrt{455}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{455}}{2*1}=\frac{-40-2\sqrt{455}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{455}}{2*1}=\frac{-40+2\sqrt{455}}{2}
$