X2+40x-1200=0

Solution for X2+40x-1200=0 equation:

X2+40X-1200=0

We add all the numbers together, and all the variables

X^2+40X-1200=0

a = 1; b = 40; c = -1200;
Δ = b2-4ac
Δ = 402-4·1·(-1200)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-80}{2*1}=\frac{-120}{2}
=-60
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+80}{2*1}=\frac{40}{2}
=20
$