p^2-16p-67=-10

Solution for p^2-16p-67=-10 equation:

p^2-16p-67=-10

We move all terms to the left:

p^2-16p-67-(-10)=0

We add all the numbers together, and all the variables

p^2-16p-57=0

a = 1; b = -16; c = -57;
Δ = b2-4ac
Δ = -162-4·1·(-57)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-22}{2*1}=\frac{-6}{2}
=-3
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+22}{2*1}=\frac{38}{2}
=19
$