30p^2-245p+40=0

Solution for 30p^2-245p+40=0 equation:

30p^2-245p+40=0

a = 30; b = -245; c = +40;
Δ = b2-4ac
Δ = -2452-4·30·40
Δ = 55225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{55225}=235$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-245)-235}{2*30}=\frac{10}{60}
=1/6
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-245)+235}{2*30}=\frac{480}{60}
=8
$