4p^2+32p+56=8

Solution for 4p^2+32p+56=8 equation:

4p^2+32p+56=8

We move all terms to the left:

4p^2+32p+56-(8)=0

We add all the numbers together, and all the variables

4p^2+32p+48=0

a = 4; b = 32; c = +48;
Δ = b2-4ac
Δ = 322-4·4·48
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*4}=\frac{-48}{8}
=-6
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*4}=\frac{-16}{8}
=-2
$