64p2+64p+12=0

Solution for 64p2+64p+12=0 equation:

64p^2+64p+12=0

a = 64; b = 64; c = +12;
Δ = b2-4ac
Δ = 642-4·64·12
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-32}{2*64}=\frac{-96}{128}
=-3/4
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+32}{2*64}=\frac{-32}{128}
=-1/4
$