2k^2-10k+3=0

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

2k^2-10k+3=0

a = 2; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·2·3
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{19}}{2*2}=\frac{10-2\sqrt{19}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{19}}{2*2}=\frac{10+2\sqrt{19}}{4}
$