How do I solve this equation?
Algebra

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(30/3x6)+(1/x2)=11/3
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First eliminate the denominator on the left hand side by multiplying them on both side, to obtain
30(x2)+1(3x6)=11/3(3x6)*(x2)
open the brackets by multiplication.ythis gives you,
30x30+3x6=11/3(3x^26x6x+12
collect the like terms together and solve them.
30x+3x306=11/3(3x^2+12)
simplify the right hand equation by dividing all through by 3
33x36=11(x^2+4)
33x36=11x^2+44
write the equation to solve.
11x^2+4433x+36=0
using quadratic fomula
x=b+ sqaureroot(b^24ac)/2a
x=33+ sqrrooot(33^24*11*80)/2*11
x=
Please let me know if you need any clarification. I'm always happy to answer your questions.kindly accept my updated corrections of the question as I realized there was an error in my working and I will be glad if you accept please.
30/3x6 + 1/x2=11/3
eliminate the denominator on the left hand side by multiplying them on both side, to obtain
30(x2)+1(3x6)=11/3(3x6)*(x2).Open brackets by multiplication to obtain
30x60+3x6=11/3(3x^26x6x+12)
30x+3x606=11/3(3x^212x+12)
33x66=11x^244x+44
move the left hand side equation to the right hand side to obtain
(11x^2)44x33x+44+66=0
NB when a positive sign cross = it becomes negative and when negative sign cross = it becomes positive.
(11x^2)  77x+110=0, divide by 11 to simplify the equation to
(x^2)7x+10=0
you find two numbers which when multiplied give us 10 and when added give us 7,
the two numbers are 5 and 2. when multiplied they give us 10 and when added they give us 7.There we substitute them in our equation to obtain.
(x^2)  2x  5x+10=0
group the equation into two segments considering where there are terms that can be simplified further with ease as
(x^2 2x) (5x+10)=0, factor the like terms as follows
x(x2)5(x2)=0.In our new equation you find that, the terms in brackets(x2) are similar,Group the terms outside the brackets together to obtain
(x5)=0 or (x2)=0
hence value of x can be x=5 or x=2.
you can also use quadratic formula to obtain the values of X
x=(b+ squareroot (b^2 4ac))/2 or x=(b  squareroot (b^2 4ac))/2
where a=1, b=7 and c=10
x=((7)+squareroot (7^2  4*1*10))/2 or x=((7)  squareroot (7^2  4*1*10))/2
x=(7+squareroot (49 40))/2 or x=(7  squareroot (49 40))/2
x=(7+ squareroot (9) )/2 or x=(7  squareroot (9) )/2
x=(7+3)/2 or x=(73)/2
x=5 or x=2
Please accept my correction and if you require more clarification on the same feel free to conduct me and I will be glad to shade more light.
kind regards
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