12k^2+13k+1=0

Solution for 12k^2+13k+1=0 equation:

12k^2+13k+1=0

a = 12; b = 13; c = +1;
Δ = b2-4ac
Δ = 132-4·12·1
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*12}=\frac{-24}{24}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*12}=\frac{-2}{24}
=-1/12
$