3=k2

Solution for 3=k2 equation:

3=k2

We move all terms to the left:

3-(k2)=0

We add all the numbers together, and all the variables

-1k^2+3=0

a = -1; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-1)·3
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-1}=\frac{0-2\sqrt{3}}{-2}
=-\frac{2\sqrt{3}}{-2}
=-\frac{\sqrt{3}}{-1}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-1}=\frac{0+2\sqrt{3}}{-2}
=\frac{2\sqrt{3}}{-2}
=\frac{\sqrt{3}}{-1}
$