3k^2+8k+2=0

Solution for 3k^2+8k+2=0 equation:

3k^2+8k+2=0

a = 3; b = 8; c = +2;
Δ = b2-4ac
Δ = 82-4·3·2
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{-(8)-2\sqrt{10}}{2*3}=\frac{-8-2\sqrt{10}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{10}}{2*3}=\frac{-8+2\sqrt{10}}{6}
$