Quadratic equation $3x^2 + x - 2 = 0$ I have $3x^2 + x - 2 = 0$ and the answers are supposed to be $-1$ and $2/3$. It was in the quadratic formula chapter so I tried to use that but since the middle x is only 1 for a coefficient, it ends up a negative number in the discriminant which isn't right for the answers. How do you get to those answers?
 A: $\rm{Discriminant} = (1)^2-4(3)(-2)=25>0$ which is actually a square!!
A: The fact that the answers are rational numbers means that the quadratic equation isn't necessary, instead you can factor.  Multiply the first and last number: $3\cdot -2 = -6$ Look for factors of $-6$ which add to give the middle term: that's easy: $+3\cdot-2 = -6$ and $+3 + -2 = +1$.  So rewrite the middle term as the sum of those two, then factor:
$$
3x^2 + 3x - 2x - 2 = 3x(x + 1) - 2(x + 1) = (3x - 2)(x + 1)
$$
Now set each factor to zero:
$$
3x - 2 = 0 \rightarrow x = \frac{2}{3}\\
x + 1 = 0 \rightarrow x = -1
$$
So the two values of $x$ which solve the equation are $x = \frac{2}{3}$ or $x = -1$.
Why does this factoring always work?
You can easily see that this type of factoring always works (when you actually can find factors which add to give the middle term) because of what you are searching for.  It all comes down to solving a "quadratic", i.e. $x^2 = b$.  How can you solve (without a calculator)?  You search for integers that squared give $b$, for instance $x^2 = 4\rightarrow x = 2$, $x^2 = 36 \rightarrow x = 6$, etc.  The difference here is that you are not searching for a perfect square but rather a "rectangle" (of sorts).  Assuming everything is an integer, you get something like the following (this is true of irrational coefficients as well, but the search space is not easily searchable):
$$
(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd
$$
So when you multiply the "first" and "last" coefficient you are multiplying $ac$ by $bd$ which gives $abcd$ which can be factored into $ad\cdot bc$ and therefore the two factors add to give $ad + bc$ which is the coefficient of the $x$-term.  Further we can see that you can factor those two to get back the original, factored, expression:
$$
acx^2 + adx + bcx + bd = ax(cx + d) + b(cx + d) = (ax + b)(cx + d) \\
acx^2 + bcx + adx + bd = cx(ax + b) + d(ax + b) = (cx + d)(ax + b)
$$
A: You have to expand the linear term. Since the coefficient of the leading term is not 1, you must multiply the coefficient of the leading term to the constant so that you see that you must have two numbers that multiply to -6 and then add to be +1. Then, you expand the linear term. For example: If n, m are you numbers you write the equation as such 
$$ 3x^2 + mx +nx -2 $$ Then you factor by grouping. Do you know how to do that? 
A: This is an alternative way of completing the square that avoids working with fractions until the last step. Instead of trying to make the $x^2$ coefficient $1$, just try making it a perfect square that is divisible by $4=2^2$.
$$\begin{array}{lll}
3x^2+x-2&=&0\\
3x^2+x&=&2\\
12(3x^2+x)&=&12\cdot 2\\
36x^2+12x&=&24\\
36x^2+12x+1&=&24+1\\
\dots
\end{array}$$
I'll let you finish.
