3k^2-14k+13=0

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

3k^2-14k+13=0

a = 3; b = -14; c = +13;
Δ = b2-4ac
Δ = -142-4·3·13
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{10}}{2*3}=\frac{14-2\sqrt{10}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{10}}{2*3}=\frac{14+2\sqrt{10}}{6}
$