40x+500=240x-3x^2

Solution for 40x+500=240x-3x^2 equation:

40x+500=240x-3x^2

We move all terms to the left:

40x+500-(240x-3x^2)=0

We get rid of parentheses

3x^2-240x+40x+500=0

We add all the numbers together, and all the variables

3x^2-200x+500=0

a = 3; b = -200; c = +500;
Δ = b2-4ac
Δ = -2002-4·3·500
Δ = 34000
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:

$\sqrt{\Delta}=\sqrt{34000}=\sqrt{400*85}=\sqrt{400}*\sqrt{85}=20\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-20\sqrt{85}}{2*3}=\frac{200-20\sqrt{85}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+20\sqrt{85}}{2*3}=\frac{200+20\sqrt{85}}{6}
$