0=-3x2+340x-3000

Solution for 0=-3x2+340x-3000 equation:

0=-3x^2+340x-3000

We move all terms to the left:

0-(-3x^2+340x-3000)=0

We add all the numbers together, and all the variables

-(-3x^2+340x-3000)=0

We get rid of parentheses

3x^2-340x+3000=0

a = 3; b = -340; c = +3000;
Δ = b2-4ac
Δ = -3402-4·3·3000
Δ = 79600
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:

$\sqrt{\Delta}=\sqrt{79600}=\sqrt{400*199}=\sqrt{400}*\sqrt{199}=20\sqrt{199}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-340)-20\sqrt{199}}{2*3}=\frac{340-20\sqrt{199}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-340)+20\sqrt{199}}{2*3}=\frac{340+20\sqrt{199}}{6}
$