9p^2+24p+16=27

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

9p^2+24p+16=27

We move all terms to the left:

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

We add all the numbers together, and all the variables

9p^2+24p-11=0

a = 9; b = 24; c = -11;
Δ = b2-4ac
Δ = 242-4·9·(-11)
Δ = 972
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:

$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-18\sqrt{3}}{2*9}=\frac{-24-18\sqrt{3}}{18}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+18\sqrt{3}}{2*9}=\frac{-24+18\sqrt{3}}{18}
$