8p^2+16p-54=0

Solution for 8p^2+16p-54=0 equation:

8p^2+16p-54=0

a = 8; b = 16; c = -54;
Δ = b2-4ac
Δ = 162-4·8·(-54)
Δ = 1984
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:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{31}}{2*8}=\frac{-16-8\sqrt{31}}{16}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{31}}{2*8}=\frac{-16+8\sqrt{31}}{16}
$