15k^2-19k+6=0

Solution for 15k^2-19k+6=0 equation:

15k^2-19k+6=0

a = 15; b = -19; c = +6;
Δ = b2-4ac
Δ = -192-4·15·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*15}=\frac{18}{30}
=3/5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*15}=\frac{20}{30}
=2/3
$