3x^2-72x+32=0

Solution for 3x^2-72x+32=0 equation:

3x^2-72x+32=0

a = 3; b = -72; c = +32;
Δ = b2-4ac
Δ = -722-4·3·32
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-40\sqrt{3}}{2*3}=\frac{72-40\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+40\sqrt{3}}{2*3}=\frac{72+40\sqrt{3}}{6}
$