3x^2-400x+600=0

Solution for 3x^2-400x+600=0 equation:

3x^2-400x+600=0

a = 3; b = -400; c = +600;
Δ = b2-4ac
Δ = -4002-4·3·600
Δ = 152800
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:

$\sqrt{\Delta}=\sqrt{152800}=\sqrt{400*382}=\sqrt{400}*\sqrt{382}=20\sqrt{382}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-400)-20\sqrt{382}}{2*3}=\frac{400-20\sqrt{382}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-400)+20\sqrt{382}}{2*3}=\frac{400+20\sqrt{382}}{6}
$