i've got $\displaystyle 3x^2-6xL+2L^2 = 0$
I want to get $\displaystyle x = 0.423L$
How do I get to there? Clear explanation needed.. ty
Assuming L is a constant you can use the quadratic formula. Because L is a constant it can be incorporated into $\displaystyle ax^2+bx+c=0$
In your case:
$\displaystyle a = 3$
$\displaystyle b = -6L$
$\displaystyle c = 2L^2$
The plug those into the quadratic formula to find x in terms of L:
$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{6L \pm \sqrt{36L^2 - 24L^2}}{6}$
simplify for L
One of the solutions is $\displaystyle \frac{3-\sqrt3}{3} L = 0.423L \text { (3dp) }$
There will be another solution though because of the plus/minus