0=3x^2-24x+28

Solution for 0=3x^2-24x+28 equation:

0=3x^2-24x+28

We move all terms to the left:

0-(3x^2-24x+28)=0

We add all the numbers together, and all the variables

-(3x^2-24x+28)=0

We get rid of parentheses

-3x^2+24x-28=0

a = -3; b = 24; c = -28;
Δ = b2-4ac
Δ = 242-4·(-3)·(-28)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{15}}{2*-3}=\frac{-24-4\sqrt{15}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{15}}{2*-3}=\frac{-24+4\sqrt{15}}{-6}
$