Solve the equation: 2x^2 + 3x = 14.

This is a quadratic equation, for these types of questions it is often easiest to try and rearrange the formula into the form ax^2 + bx + c = 0, where a, b and c are whole numbers. Therefore the first thing to do would be to take 14 away from both sides of the equation. This would leave us with 2x^2 + 3x - 14 = 0. Now that it is in the form that we want, we can start to factorise the equation (put into brackets). I find it easiest if we focus firstly on 2x^2, the only two whole numbers that can multiply together to make 2x^2 are 2x and x (as 2 times x times x is 2x^2). This means we know what the brackets start with, (2x...) and (x...).
The next step would be to focus on the last number in the equation, this would be -14. We can then write the numbers that multiply to make -14. -14 and 1, -7 and 2, -2 and 7, -1 and 14. Looking at what we have in the brackets, we know that using a 14 in the equation would not produce a 3x, therefore we are left with (7 and -2) and (2 and -7). If we then try out these number in different combinations in our brackets, we will find that the only combination that produces a +3x is (2x +7)(x-2). (As (-2)2 = -4 and 7 times 1 = 7 and then 7+(-4) = 3. Therefore the factorised equation is (2x + 7)(x - 2) = 0. If this equals 0, one of the brackets must equal 0 (as 0 multiplied by anything is still 0). If x-2=0 then x=2, alternatively if 2x+7=0 then x=-3.5, therefore x = 2 or -3.5