0=3r^2-305r+600

Solution for 0=3r^2-305r+600 equation:

0=3r^2-305r+600

We move all terms to the left:

0-(3r^2-305r+600)=0

We add all the numbers together, and all the variables

-(3r^2-305r+600)=0

We get rid of parentheses

-3r^2+305r-600=0

a = -3; b = 305; c = -600;
Δ = b2-4ac
Δ = 3052-4·(-3)·(-600)
Δ = 85825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{85825}=\sqrt{25*3433}=\sqrt{25}*\sqrt{3433}=5\sqrt{3433}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(305)-5\sqrt{3433}}{2*-3}=\frac{-305-5\sqrt{3433}}{-6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(305)+5\sqrt{3433}}{2*-3}=\frac{-305+5\sqrt{3433}}{-6}
$