16p^2+24p+9=27

Solution for 16p^2+24p+9=27 equation:

16p^2+24p+9=27

We move all terms to the left:

16p^2+24p+9-(27)=0

We add all the numbers together, and all the variables

16p^2+24p-18=0

a = 16; b = 24; c = -18;
Δ = b2-4ac
Δ = 242-4·16·(-18)
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24\sqrt{3}}{2*16}=\frac{-24-24\sqrt{3}}{32}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24\sqrt{3}}{2*16}=\frac{-24+24\sqrt{3}}{32}
$