9k^2-4=68

Solution for 9k^2-4=68 equation:

9k^2-4=68

We move all terms to the left:

9k^2-4-(68)=0

We add all the numbers together, and all the variables

9k^2-72=0

a = 9; b = 0; c = -72;
Δ = b2-4ac
Δ = 02-4·9·(-72)
Δ = 2592
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:

$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{2}}{2*9}=\frac{0-36\sqrt{2}}{18}
=-\frac{36\sqrt{2}}{18}
=-2\sqrt{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{2}}{2*9}=\frac{0+36\sqrt{2}}{18}
=\frac{36\sqrt{2}}{18}
=2\sqrt{2}
$