0.5x^2+40x+(-300)=0

Solution for 0.5x^2+40x+(-300)=0 equation:

0.5x^2+40x+(-300)=0

We add all the numbers together, and all the variables

0.5x^2+40x-300=0

a = 0.5; b = 40; c = -300;
Δ = b2-4ac
Δ = 402-4·0.5·(-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*0.5}=\frac{-40-10\sqrt{22}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{22}}{2*0.5}=\frac{-40+10\sqrt{22}}{1}
$