300=80x-0.5x^2-(40x+300)

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

300=80x-0.5x^2-(40x+300)

We move all terms to the left:

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


We calculate terms in parentheses: -(80x-0.5x^2-(40x+300)), so:

80x-0.5x^2-(40x+300)

determiningTheFunctionDomain
-0.5x^2+80x-(40x+300)

We get rid of parentheses

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

We add all the numbers together, and all the variables

-0.5x^2+40x-300

Back to the equation:

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


We get rid of parentheses

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

We add all the numbers together, and all the variables

0.5x^2-40x+600=0

a = 0.5; b = -40; c = +600;
Δ = b2-4ac
Δ = -402-4·0.5·600
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20}{2*0.5}=\frac{20}{1}
=20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20}{2*0.5}=\frac{60}{1}
=60
$