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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -((-0.5x^2+40x-300)), so:

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

We get rid of parentheses

-0.5x^2+40x-300

Back to the equation:

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


We get rid of parentheses

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