3k^2+19k+20=0

Solution for 3k^2+19k+20=0 equation:

3k^2+19k+20=0

a = 3; b = 19; c = +20;
Δ = b2-4ac
Δ = 192-4·3·20
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*3}=\frac{-30}{6}
=-5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*3}=\frac{-8}{6}
=-1+1/3
$