-0.5x^2+40x-300=50

Solution for -0.5x^2+40x-300=50 equation:

-0.5x^2+40x-300=50

We move all terms to the left:

-0.5x^2+40x-300-(50)=0

We add all the numbers together, and all the variables

-0.5x^2+40x-350=0

a = -0.5; b = 40; c = -350;
Δ = b2-4ac
Δ = 402-4·(-0.5)·(-350)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-30}{2*-0.5}=\frac{-70}{-1}
=+70
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+30}{2*-0.5}=\frac{-10}{-1}
=+10
$