k2=32

Solution for k2=32 equation:

k2=32

We move all terms to the left:

k2-(32)=0

We add all the numbers together, and all the variables

k^2-32=0

a = 1; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·1·(-32)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*1}=\frac{0-8\sqrt{2}}{2}
=-\frac{8\sqrt{2}}{2}
=-4\sqrt{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*1}=\frac{0+8\sqrt{2}}{2}
=\frac{8\sqrt{2}}{2}
=4\sqrt{2}
$