16p^2+4p=20

Solution for 16p^2+4p=20 equation:

16p^2+4p=20

We move all terms to the left:

16p^2+4p-(20)=0

a = 16; b = 4; c = -20;
Δ = b2-4ac
Δ = 42-4·16·(-20)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*16}=\frac{-40}{32}
=-1+1/4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*16}=\frac{32}{32}
=1
$