41p-28-15p^2=0

Solution for 41p-28-15p^2=0 equation:

41p-28-15p^2=0

a = -15; b = 41; c = -28;
Δ = b2-4ac
Δ = 412-4·(-15)·(-28)
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{-(41)-1}{2*-15}=\frac{-42}{-30}
=1+2/5
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+1}{2*-15}=\frac{-40}{-30}
=1+1/3
$