3x^2-48x+120=0

Solution for 3x^2-48x+120=0 equation:

3x^2-48x+120=0

a = 3; b = -48; c = +120;
Δ = b2-4ac
Δ = -482-4·3·120
Δ = 864
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:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12\sqrt{6}}{2*3}=\frac{48-12\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12\sqrt{6}}{2*3}=\frac{48+12\sqrt{6}}{6}
$