14z^2+13z+3=0

Solution for 14z^2+13z+3=0 equation:

14z^2+13z+3=0

a = 14; b = 13; c = +3;
Δ = b2-4ac
Δ = 132-4·14·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*14}=\frac{-14}{28}
=-1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*14}=\frac{-12}{28}
=-3/7
$