3k2=12

Solution for 3k2=12 equation:

3k^2=12

We move all terms to the left:

3k^2-(12)=0

a = 3; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·3·(-12)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*3}=\frac{-12}{6}
=-2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*3}=\frac{12}{6}
=2
$