p^2-16p-50=0

Solution for p^2-16p-50=0 equation:

p^2-16p-50=0

a = 1; b = -16; c = -50;
Δ = b2-4ac
Δ = -162-4·1·(-50)
Δ = 456
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:

$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{114}}{2*1}=\frac{16-2\sqrt{114}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{114}}{2*1}=\frac{16+2\sqrt{114}}{2}
$