6p^2-48p+40=0

Solution for 6p^2-48p+40=0 equation:

6p^2-48p+40=0

a = 6; b = -48; c = +40;
Δ = b2-4ac
Δ = -482-4·6·40
Δ = 1344
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{21}}{2*6}=\frac{48-8\sqrt{21}}{12}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{21}}{2*6}=\frac{48+8\sqrt{21}}{12}
$