2=21l2

Solution for 2=21l2 equation:

2=21l^2

We move all terms to the left:

2-(21l^2)=0

a = -21; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-21)·2
Δ = 168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*-21}=\frac{0-2\sqrt{42}}{-42}
=-\frac{2\sqrt{42}}{-42}
=-\frac{\sqrt{42}}{-21}
$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*-21}=\frac{0+2\sqrt{42}}{-42}
=\frac{2\sqrt{42}}{-42}
=\frac{\sqrt{42}}{-21}
$