6p^2-31p+40=0

Solution for 6p^2-31p+40=0 equation:

6p^2-31p+40=0

a = 6; b = -31; c = +40;
Δ = b2-4ac
Δ = -312-4·6·40
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-1}{2*6}=\frac{30}{12}
=2+1/2
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+1}{2*6}=\frac{32}{12}
=2+2/3
$