2p^2+26p+6p=48

Solution for 2p^2+26p+6p=48 equation:

2p^2+26p+6p=48

We move all terms to the left:

2p^2+26p+6p-(48)=0

We add all the numbers together, and all the variables

2p^2+32p-48=0

a = 2; b = 32; c = -48;
Δ = b2-4ac
Δ = 322-4·2·(-48)
Δ = 1408
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{22}}{2*2}=\frac{-32-8\sqrt{22}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{22}}{2*2}=\frac{-32+8\sqrt{22}}{4}
$