Question 23410
{{{10c^2-21c=-4c+6}}}


Are you sure you didn't copy this one wrong?  It works out a LOT better (it factors!) if you had written this {{{10c^2 -21c = -4c -6}}}.  I'm going to take the liberty of changing it, and if I'm wrong, then just repost the question.  I will owe you a solution to the other problem, using the quadratic formula (or one of algebra.com quadratic equation solver!)


So I'm saying let's do this: 
{{{10c^2 -21c = -4c -6}}}.  


This is a quadratic equation, so set it equal to zero, by adding +4c +6 to each side:


{{{10c^2 -21c +4c +6 = -4c + 6 +4c+6 }}}
{{{10c^2 -17c +6 = 0}}}


This can be a tough factoring problem.  You might want to see a page on my website:  click on my tutor name "rapaljer" anywhere in algebra.com, and look for "Basic Algebra", then go to "Samples of Basic Algebra: One Step at a Time."  Then in Chapter 2 there are several sections on factoring.  The one you need here for this problem is the one I call "Advanced Trinomial Factoring."  It is a DETAILED explanation, with examples, exercises, and solutions to ALL the exercises-- and like everything else around here, it's ALL FREE!!


Back to the problem:

{{{10c^2 -17c +6 = 0}}}

You need to find two numbers whose product is 10c^2.  Most likely that will be 5c*2c:

{{{(5c______)*(2c______) = 0}}}


Now, you need to find two numbers whose product is +6, and the middle term has to add up to a  -17c.  The signs will both be negative, so try -6*-1 (since I happen to have already tried the -3*-2 and it didn't work out right.  I think you should putting the -6 first and the -1 in the second slot.

{{{(5c-6)(2c-1)= 0}}}

If you will do the OUTER times OUTER and the INNER times INNER, you will have -5c -12c, which is -17c.


Now finish the problem by separating the quadratic equation into TWO solutions:

{{{(5c-6)(2c-1)= 0}}}


First solution:
{{{5c-6=0}}}
{{{5c=6}}}
{{{c= 6/5}}}


Second solution:

{{{2c-1= 0}}}
{{{2c=1}}}
{{{c= 1/2}}}


R^2 at SCC