# Solve 3x^2 = 8x - 2 giving your answers to 2 d.p.

First things first, we have to decide what **type** of equation this is, so we can choose how to deal with it. If it were linear (which means every term involving an 'x' has no power/superscript on it. I.e. x^{2 }or x^{3} do not appear in your equation), then we could simply rearrange to solve. But our equation has a 3x^{2} term, so we need to do something else.

First, let's put everything on one side, so we can recognise that this equation is actually a **quadratic equation**: subtracting 8x and adding 2 to both sides gives **3x ^{2} - 8x + 2 = 0**.

Now, there are 3 ways we can deal with a quadratic equation: **Factorisation**, **Completeing the Square, **and the** Quadratic Formula. **

The first two methods won't really help us here, but how did I figure that out? Well, the long way would be to do some trial and error, attempting to factorise. You would quickly see you can't make any combination work. When factorising fails, you normally try the quadratic formula next. **Here's a tip**: if the question asks you to give your answer to a certain number of decimal places, it means you cannot factorise the quadratic because the answer is a fraction or surd, so **always use the quadratic formula in this case!**

So now we know we're going to apply the quadratic formula to this equation. Here is the formula in words (because I can't format it clearly on here), with brackets added to show you in what order to do operations:

x = [(-b) plus or minus the (square root of {b^{2} - 4ac})] all over 2a

(Remember to **divide everything** by the 2a! It is a common mistake not to do this).

We have 3 variables, a,b and c, which we can find from our quadratic equation we are trying to solve. **Note**: our quadratic **MUST** be in the form ax^{2}+bx+c=0 for this to work. We did this earlier, so we can proceed.

From our equation, a (the bit in front of the x^{2}) equals +3. b (the bit in front of the x) equals -8. (The minus is **important!!**). c (the number without an x) equals +2.

Now we can put a=3, b=-8, c=2 into the quadratic equation.

Note this **common mistake**: If b is negative, like we have in this question, what happens? -b, here, equals -(-8), and two minuses make a plus, so this becomes **+**8.

For the -b^{2}, it is important to know where the brackets should be. This is the same thing as (-b)^{2} and therefore is **ALWAYS positive, **because all square numbers are positive numbers. So in our question, (-b)^{2} is (-8)^{2}=64.

4ac = 4*3*2 = 24. Therefore, inside the square root we have 64-24=40. 2a = 2*3 = 6.

Now we can put (8 plus/minus the square root of 40)/6 into our calculator, to give us two answers (one for the plus and one for the minus). These are 2.3874... and 0.2792...

Rounding these to 2 decimal places gives us our final solutions to our quadratic: 2.39 and 0.28.