3k^2+19k+28=0

Solution for 3k^2+19k+28=0 equation:

3k^2+19k+28=0

a = 3; b = 19; c = +28;
Δ = b2-4ac
Δ = 192-4·3·28
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*3}=\frac{-24}{6}
=-4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*3}=\frac{-14}{6}
=-2+1/3
$