0=3r^2-28r+48

Solution for 0=3r^2-28r+48 equation:

0=3r^2-28r+48

We move all terms to the left:

0-(3r^2-28r+48)=0

We add all the numbers together, and all the variables

-(3r^2-28r+48)=0

We get rid of parentheses

-3r^2+28r-48=0

a = -3; b = 28; c = -48;
Δ = b2-4ac
Δ = 282-4·(-3)·(-48)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{13}}{2*-3}=\frac{-28-4\sqrt{13}}{-6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{13}}{2*-3}=\frac{-28+4\sqrt{13}}{-6}
$