p^2-26p-169=0

Solution for p^2-26p-169=0 equation:

p^2-26p-169=0

a = 1; b = -26; c = -169;
Δ = b2-4ac
Δ = -262-4·1·(-169)
Δ = 1352
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:

$\sqrt{\Delta}=\sqrt{1352}=\sqrt{676*2}=\sqrt{676}*\sqrt{2}=26\sqrt{2}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-26\sqrt{2}}{2*1}=\frac{26-26\sqrt{2}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+26\sqrt{2}}{2*1}=\frac{26+26\sqrt{2}}{2}
$