3x^2-48x+36=0

Solution for 3x^2-48x+36=0 equation:

3x^2-48x+36=0

a = 3; b = -48; c = +36;
Δ = b2-4ac
Δ = -482-4·3·36
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12\sqrt{13}}{2*3}=\frac{48-12\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12\sqrt{13}}{2*3}=\frac{48+12\sqrt{13}}{6}
$