6+k^2=96

Solution for 6+k^2=96 equation:

6+k^2=96

We move all terms to the left:

6+k^2-(96)=0

We add all the numbers together, and all the variables

k^2-90=0

a = 1; b = 0; c = -90;
Δ = b2-4ac
Δ = 02-4·1·(-90)
Δ = 360
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:

$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*1}=\frac{0-6\sqrt{10}}{2}
=-\frac{6\sqrt{10}}{2}
=-3\sqrt{10}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*1}=\frac{0+6\sqrt{10}}{2}
=\frac{6\sqrt{10}}{2}
=3\sqrt{10}
$