169z2+13z-1=0

Solution for 169z2+13z-1=0 equation:

169z^2+13z-1=0

a = 169; b = 13; c = -1;
Δ = b2-4ac
Δ = 132-4·169·(-1)
Δ = 845
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{845}=\sqrt{169*5}=\sqrt{169}*\sqrt{5}=13\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13\sqrt{5}}{2*169}=\frac{-13-13\sqrt{5}}{338}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13\sqrt{5}}{2*169}=\frac{-13+13\sqrt{5}}{338}
$