300x=-4x^2+40x+300

Solution for 300x=-4x^2+40x+300 equation:

300x=-4x^2+40x+300

We move all terms to the left:

300x-(-4x^2+40x+300)=0

We get rid of parentheses

4x^2-40x+300x-300=0

We add all the numbers together, and all the variables

4x^2+260x-300=0

a = 4; b = 260; c = -300;
Δ = b2-4ac
Δ = 2602-4·4·(-300)
Δ = 72400
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:

$\sqrt{\Delta}=\sqrt{72400}=\sqrt{400*181}=\sqrt{400}*\sqrt{181}=20\sqrt{181}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(260)-20\sqrt{181}}{2*4}=\frac{-260-20\sqrt{181}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(260)+20\sqrt{181}}{2*4}=\frac{-260+20\sqrt{181}}{8}
$