3k^2-13k+10=0

Solution for 3k^2-13k+10=0 equation:

3k^2-13k+10=0

a = 3; b = -13; c = +10;
Δ = b2-4ac
Δ = -132-4·3·10
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-7}{2*3}=\frac{6}{6}
=1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+7}{2*3}=\frac{20}{6}
=3+1/3
$