3k^2+8k=19

Solution for 3k^2+8k=19 equation:

3k^2+8k=19

We move all terms to the left:

3k^2+8k-(19)=0

a = 3; b = 8; c = -19;
Δ = b2-4ac
Δ = 82-4·3·(-19)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{73}}{2*3}=\frac{-8-2\sqrt{73}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{73}}{2*3}=\frac{-8+2\sqrt{73}}{6}
$