3x^2-240x+4600=0

Solution for 3x^2-240x+4600=0 equation:

3x^2-240x+4600=0

a = 3; b = -240; c = +4600;
Δ = b2-4ac
Δ = -2402-4·3·4600
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-20\sqrt{6}}{2*3}=\frac{240-20\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+20\sqrt{6}}{2*3}=\frac{240+20\sqrt{6}}{6}
$