p^2+16p-4=0

Solution for p^2+16p-4=0 equation:

p^2+16p-4=0

a = 1; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·1·(-4)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{17}}{2*1}=\frac{-16-4\sqrt{17}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{17}}{2*1}=\frac{-16+4\sqrt{17}}{2}
$