-0.5x^2+80x-40x-300=0

Solution for -0.5x^2+80x-40x-300=0 equation:

-0.5x^2+80x-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 = 1000
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:

$\sqrt{\Delta}=\sqrt{1000}=\sqrt{100*10}=\sqrt{100}*\sqrt{10}=10\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{10}}{2*-0.5}=\frac{-40-10\sqrt{10}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{10}}{2*-0.5}=\frac{-40+10\sqrt{10}}{-1}
$