p^2=169p

Solution for p^2=169p equation:

p^2=169p

We move all terms to the left:

p^2-(169p)=0

a = 1; b = -169; c = 0;
Δ = b2-4ac
Δ = -1692-4·1·0
Δ = 28561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{28561}=169$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-169)-169}{2*1}=\frac{0}{2}
=0
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-169)+169}{2*1}=\frac{338}{2}
=169
$