9k^2+66k=21

Solution for 9k^2+66k=21 equation:

9k^2+66k=21

We move all terms to the left:

9k^2+66k-(21)=0

a = 9; b = 66; c = -21;
Δ = b2-4ac
Δ = 662-4·9·(-21)
Δ = 5112
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:

$\sqrt{\Delta}=\sqrt{5112}=\sqrt{36*142}=\sqrt{36}*\sqrt{142}=6\sqrt{142}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(66)-6\sqrt{142}}{2*9}=\frac{-66-6\sqrt{142}}{18}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(66)+6\sqrt{142}}{2*9}=\frac{-66+6\sqrt{142}}{18}
$