3k2+22k=16

Solution for 3k2+22k=16 equation:

3k^2+22k=16

We move all terms to the left:

3k^2+22k-(16)=0

a = 3; b = 22; c = -16;
Δ = b2-4ac
Δ = 222-4·3·(-16)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*3}=\frac{-48}{6}
=-8
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*3}=\frac{4}{6}
=2/3
$