Find the possible values of x from the equation 3(x^2)+2x-4=2(x^2)+3x+8

Here we have an example of a quadratic equation in terms of x. Start by manipulating the equation until it is equal to 0: subtract 2(x^2) from both sides to get (x^2)+2x-4=3x+8. Then subtract 3x from both sides to get (x^2)-x-4=8. Next subtract 8 from both sides to get (x^2)-x-12=0.
Now we must factorise the equation. Write out two sets of brackets. As there is a singular (x^2) in the equation, the first term in each bracket will be an x, as these would multiply together to get the (x^2) term if you were to multiply out the brackets; this is the logic for all of the factorisation, as the aim here is simply to represent the equation in an alternate form while still making sure to keep the value of it the same. This should look like this: (x.....)(x......)=0. Then, as the -x term has a magnitude of 1, you must think of factors of 12 with a difference of 1. Here, this is 3 and 4. As the constant term here (12) is negative, one of the 3 and the 4 must be negative and one positive as then when they are multiplied together they give a negative result (-12). We have now reached a stage where you put the numbers selected into your brackets like this: (x....4)(x....3). You decide which one will take which sign by how many x's you have in the equation equal to 0: here, we have -x (read minus one x) and 3-4=-1, so the 4 takes the negative sign. Hence, you put these signs in to get (x-4)(x+3)=0. Given that this equation is equal to 0, at least one of the brackets alone must be equal to 0: hence, you can say that (x-4)=0 or (x+3)=0, and solving these equations gives the result x=4 or x=-3. You can check these results by substituting them back into any equation stated before the factorisation (brackets) stage.