Question 693410
For quadratics you need everything on the left side of the equal sign:

{{{3u^2-5u-2=0}}}

There are 2 ways of solving quadratics. 

A. You try to find to numbers that their product is equal to the product of the u^2 coefficient and the number without letter (in this case {{{3(-2)=-6}}}), and their sum is equal to the u coefficient (in this case {{{-5}}}). The numbers that multiplied give {{{-6}}} and added give {{{-5}}} are : {{{-6}}} and {{{+1}}}.

Instead of {{{-5u}}} in the original equation write the two numbers multiplied by {{{u}}}:

{{{3u^2 -6u+u-2=0}}}

Group the four terms in 2 pairs: first two terms and last two terms.

{{{(3u^2-6u)+(u-2)=0}}}

See if you can factor out anything from those parentheses (you always have at least the letter from the first parenthesis):

{{{3u(u-2)+1(u-2)=0}}}

If you done everything right the two parentheses are the same at this point. This repeated parenthesis is one of the 2 parentheses you want to solve the quadratic. The second one is put together by what you factored out:

{{{(u-2)(3u+1)=0}}}

For the quadratic solutions take each parenthesis and set it equal to zero.

First solution: 

{{{u-2=0}}}
{{{u=2}}}

Second solution:

{{{3u+1=0}}}
{{{u=-1/3}}}

This method does not always work. 

B. The second method is to know the quadratic formula, and works every time:

If {{{ax^2+bx+c=0}}} then the solutions are given by:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

In your case {{{3u^2-5u-2=0}}}, {{{a=3}}}, {{{b=-5}}}, {{{c=-2}}}.

{{{u=(5 +- sqrt( 25-4*(3)*(-2) ))/6 }}}

{{{u=(5 +- sqrt( 25+24))/6 }}}

{{{u=(5 +- sqrt( 49))/6 }}}

{{{u=(5 +- 7)/6 }}}

First solution:

{{{u = (5+7)/6 = 12/6=2}}}

Second solution:

{{{u=(5-7)/6=-2/6=-1/3}}}