p^2+16p-42=64

Solution for p^2+16p-42=64 equation:

p^2+16p-42=64

We move all terms to the left:

p^2+16p-42-(64)=0

We add all the numbers together, and all the variables

p^2+16p-106=0

a = 1; b = 16; c = -106;
Δ = b2-4ac
Δ = 162-4·1·(-106)
Δ = 680
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:

$\sqrt{\Delta}=\sqrt{680}=\sqrt{4*170}=\sqrt{4}*\sqrt{170}=2\sqrt{170}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{170}}{2*1}=\frac{-16-2\sqrt{170}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{170}}{2*1}=\frac{-16+2\sqrt{170}}{2}
$