2x^2 - 12 = 2x
I can’t figure it out.
That depends on what your goal is. If you just need the real roots, for whatever reason, you use a computer tool, like typing it into WolframAlpha:
Why are you trying to solve it and what mathematical knowledge do you already possess?
I’m trying to solve it like a quadratic equation.
That’s smart, since it is a quadratic equation, but there are lots of ways to solve them.
I’m going to hazard a guess you’re trying to use the quadratic formula and the 2x being on the right side of the equal sign is what confuses you?
Move it to the same side as the rest so you get: 2x^2 - 2x - 12 = 0
Then divide through by 2 and you will find you don’t need the quadratic formula - the expression factorises.
As the quadratic formula shows, a quadratic equation (formula) has two solutions (or roots or zeros). These are usually distinct, but they might be the same as well (e.g., for the equation x^2=0, both the roots are equal to 0). Moreover, the solutions might be integers, rational numbers, real numbers, or even complex numbers.
The first thing to try before hitting it with the quadratic formula is to look for integer solutions. In this particular case, the equation can be divided by 2, to get x^2-x-6=0.
Then you try some simple numbers to see if they are solutions to the equation. Does x=0 satisfy the equation? Let’s try: substituting x=0 in the left side gives 0^2-0-6 = -6, not zero. Then do the same with 1: is 1^2-1-6 equal to zero? (Answer: no). You do this for some initial values. For this example, you will quickly discover one of the roots this way (hint: it’s a single digit integer).
Now, there’s no need for the solutions to be nonnegative. What about negative numbers? So, let’s try those as well. x=-1, -2, -3, … You could do the same thing with the negative numbers. The procedure is the same; just be careful not to get tripped up over the minus signs. For example, for x=-1, the left hand side evaluates to:
(-1)^2 - (-1) - 6 = +1 +1-6 = …
which is not zero, so -1 is not a solution. You could try further to see if you could discover the second solution. (Hint: The second solution is a small negative integer.)
This above method is for quickly guessing solutions for easy equations. If you want something that works in all cases, the quadratic formula is the way to go, except it’s slightly calculation heavy (and has more chances for mistakes). If the equation does not integer roots, or if it has but you are not able to figure them out, then you might have to use the formula anyway.
There are more clever things that you can do (e.g., if you know one solution, you can use that to easily find the second solution), but those methods need some familiarity simple properties of quadratic equations. How familiar are you with them?
This seems like the quickest way. x^2 - x - 6 = 0 factors to (x-3)(x+2)=0
Right? We learned the FOIL method when I was in school: First, Outside, Inside, Last.
So that makes it easy: if (x-3) multiplied by (x+2) is zero, then either x-3=0 or x+2=0.
The second step is overly complicated. To figure out if there is an integer answer, first find the factors for 6, 6 and 1, 3 and 2. Since the x has a minus before it the two factors have different signs, so take the factors and see if any of them can be subtracted from the other to give you the coefficient of x, which is 1. 3-2=1 so the factors are (x-3)(x+2)
As pointed out by other helpful folks, you can only solve for roots or factorize if all of the “x” terms are on one side of the equation and a “0” is on the other. That’s simple subtraction or addition. (Subtraction, in this case.)
As also pointed out, once you have “0” on the right, you can divide the entire polynomial by a non-zero constant (both sides, but dividing 0 is a doddle) to reduce the coefficients to their lowest factor. Since, for instance, 2x[sup]2[/sup] and x[sup]2[/sup] cross zero at the same point – have the same root.
Praise point for Bricker. #mathteacher
This is kind of what the last paragraph in my earlier response refers to. 
Yes, it is inefficient to do the guessing game all over again for the second solution, when we can readily calculate it from the first solution and the coefficients of the equation. But I wasn’t sure what level of familiarity to assume from the op.
We had a thread on factoring a while back explain factoring quadratic equations by grouping to me like I'm an idiot - Factual Questions - Straight Dope Message Board by the way.
I find this a curious comment. Every quadratic expression factorizes. The question is what it factorizes into. And that’s precisely the question the quadratic formula is designed to answer.
(I mean, I know people are using “factorizes” to mean “factorizes over the integers”. It’s just curious that people sometimes act as if it is immediately obvious how to recognize such a situation and furthermore factor such a quadratic (so obvious as to need no further explanation!))
And there are deterministic algorithms to determining if a given quadratic factors over the integers and if so, what those factors are so you can move beyond intuition or guess-and-check.
One deterministic algorithm is to simply find the solutions using the quadratic formula and check to see whether they’re integers. But I suspect that’s not what you mean.
Another deterministic algorithm is to factor both A and C, and try every combination of those factors to see if you replicate the B term. This is what students usually do when they’re “solving by factoring”, and in some ways it’s easier to understand, but in the general case it turns out to be much harder than using the quadratic formula (since factoring integers is already in general a hard problem).
There are various refinements you can apply to the second method to decrease the amount of guess-and-check needed, but none of them remove the need for factoring the integers.
Everything Chronos says is right. I will re-say part of it.
Ax[sup]2[/sup] + Bx + C factors over the integers (i.e., is of the form W * (x - Y) * (x - Z) for some integers W, Y, and Z) just in case A, B, and C are all integers, B and C are both divisible by A, and B[sup]2[/sup] - 4AC (equivalently, (B/A)[sup]2[/sup] - 4C/A) is a perfect square.
In that case, the factorization is such that W = A, while Y and Z are given as (-B plus or minus sqrt(B[sup]2[/sup] - 4AC))/2A. [This is the quadratic formula]
Ah, but I suppose what people are really interested in is factorization over the integers in the sense of a decomposition as (Wx - Y) * (Zx - Q), for integers W, Y, Z, and Q.
Such factorization is possible just in case A, B, and C are integers and B[sup]2[/sup] - 4AC is a square rational (i.e., rational with perfect square numerator and denominator). Again, the quadratic formula determines the decomposition (with room to swap integer factors between the two degree 1 polynomials).
Er, what happened to my edit? I meant for that second post to be as follows:
Ah, but I suppose what people are really interested in is factorization over the integers in the sense of a decomposition as (Wx - Y) * (Zx - Q), for integers W, Y, Z, and Q.
Such factorization is possible just in case A, B, and C are integers and B[sup]2[/sup] - 4AC is a perfect square. Again, the quadratic formula determines the decomposition (with room to swap integer factors between the two degree 1 polynomials). If B and C are not both divisible by A, there will be non-integer (rational) roots, but that’s ok on this account of “factorizability”.
You can solve it numerically too. Re- arrange to x= sqrt (x+6) and x = -sqrt (x+6)
So, using the simple method X_new = sqrt(X_old +6) and X_new = -sqrt(X_old +6)
Assume X_old = 0, and iterate
for the positive equation you get :
2.449
2.907
2.984
2.997
3.000
3.000
For the negative equation, you get :
-2.449
-1.884
-2.029
-1.993
-2.002
-2.000
-2.000
Just a fun exercise - the above methods are easier. However, you can do this one on a calculator.