3k^2+20k+9=-3

Solution for 3k^2+20k+9=-3 equation:

3k^2+20k+9=-3

We move all terms to the left:

3k^2+20k+9-(-3)=0

We add all the numbers together, and all the variables

3k^2+20k+12=0

a = 3; b = 20; c = +12;
Δ = b2-4ac
Δ = 202-4·3·12
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*3}=\frac{-36}{6}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*3}=\frac{-4}{6}
=-2/3
$