How do I put this in the calculator and what is the answer please?
(((2x+x*sqrt(3))/2))(((2x+x*sqrt(3))/2-x)((2x+x*sqrt(3))/2-x))*((2x+x*sqrt(3))/2-x*sqrt(3)))=9*sqrt(3)
+118613 (((2x+x*sqrt(3))/2))(((2x+x*sqrt(3))/2-x)((2x+x*sqrt(3))/2-x))*((2x+x*sqrt(3))/2-x*sqrt(3)))=9*sqrt(3)
First I want to make sense of it!
Is this your question?
$$\left(\frac{(2x+x\sqrt3)}{2}\right)\left(\frac{(2x+x\sqrt3)}{2}-x\right)\left(\frac{(2x+x\sqrt3)}{2}-x\right)\left(\frac{(2x+x\sqrt3)}{2}-x\sqrt3\right)=9\sqrt3$$
It's too long - it won't display on one line!
+118613 I think that is the same as
$$(2x+x\sqrt3)(2x+x\sqrt3-2x)^2(2x+x\sqrt3-2x\sqrt3)=144\sqrt3\\
Do you agree?\\
(2x+x\sqrt3)(x\sqrt3)^2(2x-x\sqrt3)=144\sqrt3\\
3x^2((2x)^2-(x\sqrt3)^2)=144\sqrt3\\
3x^2(4x^2-3x^2)=144\sqrt3\\\\
3x^2(x^2)=144\sqrt3\\\\
3x^4=144\sqrt3\\\\
x^4=\frac{144\sqrt3}{3}\\\\
x=\pm\left(\frac{144\sqrt3}{3}\right)^{(1/4)}\\\\$$
$${\left({\frac{{\mathtt{144}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}{{\mathtt{3}}}}\right)}^{{\mathtt{0.25}}} = {\mathtt{3.019\: \!607\: \!296\: \!954\: \!21}}$$
I did this on the site calc
x ≈ ± 3.019607
That's assuming I didn't make any stupid mistakes.
Now you can either tell me I am wrong or you can say thank you! 
+118613 Thank you - We really appreciate it whe people say thanks. 
(You got 4 more points too - 3 from me and 1 from someone else, probably another answerer)
+128707 Yep...from me.......I always like to get feedback, too. I'm not always correct - or even precise - with my answers, but I try hard. One thing this board has taught me.......you can't just throw something imprecise out there.......but that's a good thing.....it makes you think a little more about the answers you give!!!
