.5x^2+40x-300=0

Solution for .5x^2+40x-300=0 equation:

.5x^2+40x-300=0

a = .5; b = 40; c = -300;
Δ = b2-4ac
Δ = 402-4·.5·(-300)
Δ = 2200
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:

$\sqrt{\Delta}=\sqrt{2200}=\sqrt{100*22}=\sqrt{100}*\sqrt{22}=10\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{22}}{2*.5}=\frac{-40-10\sqrt{22}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{22}}{2*.5}=\frac{-40+10\sqrt{22}}{1}
$