-0.5x^2+40x+15300=0

Solution for -0.5x^2+40x+15300=0 equation:

-0.5x^2+40x+15300=0

a = -0.5; b = 40; c = +15300;
Δ = b2-4ac
Δ = 402-4·(-0.5)·15300
Δ = 32200
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:

$\sqrt{\Delta}=\sqrt{32200}=\sqrt{100*322}=\sqrt{100}*\sqrt{322}=10\sqrt{322}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{322}}{2*-0.5}=\frac{-40-10\sqrt{322}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{322}}{2*-0.5}=\frac{-40+10\sqrt{322}}{-1}
$