Solving for x using the quadratic equation

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sportsaholic2397

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Alright so i am having a hard time towards the end of this problem i will show my work, and feel free to say if there is a spot that i screwed up, but there is one in particular spot that I can not continue on...

Solve For X

3x^2-X-5=0
a=3 b=-1 c=-5

-(-1)+/-?(-1^2-4(3)(-5))/(2(3))

(1)+/-?(1+60)/(6)

(1)+/-?(61)/(6)....now this is where i get lost. I have looked at some other homeworks, and looks like they usually simplify or get the LCD...but im not sure if that is an option...could anyone help me figure out how to finish this problem and find X?
 
sportsaholic2397 said:
nice call thanks. alright still dont know how to approach this...
Click to expand...
I don't understand what more there is to it.
You found the two roots: \(\displaystyle \frac{{1 + \sqrt {61} }}{6}\;\& \,\frac{{1 - \sqrt {61} }}{6}\).

What more do you think that you need to do?
 
haha ok well then thats that, i am just very hesitant with math because i have had my fair share of challenges with it so im just makin sure i dont get anything wrong.. but if thats it then im set. thanks for the help. just needed a 2nd set of eyes and i will use this for help with future similar problems.
 
sportsaholic2397 said:
3x^2-X-5=0
a=3 b=-1 c=-5
Click to expand...
Are you sure that's not 3x^2 - 2x - 5 = 0 ? Perhaps you made a typo?
If so, you'd get a much "nicer" solution...try it!
 
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