quadratic solution

**custer** · May 29th 2009, 04:15 AM

i've got $3x^2-6xL+2L^2 = 0$

I want to get $x = 0.423L$

How do I get to there? Clear explanation needed.. ty

**e^(i*pi)** · May 29th 2009, 04:22 AM

Originally Posted by custer

i've got $3x^2-6xL+2L^2 = 0$

I want to get $x = 0.423L$

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 $ax^2+bx+c=0$

In your case:

$a = 3$
$b = -6L$
$c = 2L^2$

The plug those into the quadratic formula to find x in terms of L:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{6L \pm \sqrt{36L^2 - 24L^2}}{6}$

simplify for L

One of the solutions is $\frac{3-\sqrt3}{3} L = 0.423L \text { (3dp) }$

There will be another solution though because of the plus/minus

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