 So let's take a look at a general approach to solving quadratic equations known as the quadratic formula. And just a quick recap in general, the slowest and least efficient way of solving equations is to try to factor them. It's usually a waste of time because outside of some very, very, very, very, very, very limited and very, very carefully constructed examples designed to show how you can solve an equation by factoring, most equations will not be factorable. So it's usually a waste of time trying to factor because it can't be done. On the other hand, completing the square is both faster and more efficient than factoring, and will always work. Now, what we can do is we can combine all of the steps that we use to complete the square and we will create what's known as the quadratic formula. Now, we'll go over the derivation in class, but for now the quadratic formula is going to be the following. If I have any quadratic equation, ax squared plus bx plus c, equal to zero, then the solutions are going to be found by the following, x equals negative b, that's our coefficient of x, plus or minus b squared, minus 4ac, all over 2a. There are extremely few things that you should memorize in mathematics. The quadratic formula is one of them. One of the things that you should memorize, that you should actually commit to memory as a formula, is the quadratic formula. And this is because it's so useful. It shows up in so many contexts that while it's not really necessary, you can solve quadratics by completing the square. It's easy enough to memorize the quadratic formula and useful enough that it's worth doing. So, memorize it. Got it? Okay, let's go on. Well, realistically you'll use it a bunch of times and then eventually you'll memorize it. So, let's take a look. Here's our quadratic equation, 3x squared plus 4x, minus 1, set equal to zero. Now, if we do want to check out a few things, we could only use the quadratic formula if we have a quadratic polynomial, a second-degree polynomial equation where we have it set equal to zero. So, let's see, 3x squared plus 4x, this is a second-degree polynomial, equal to zero. This is a quadratic equation and so we can solve it using the quadratic formula. And so, again, paper is cheap, we'll write down the quadratic formula. It helps to write it down a few times. It helps to memorize it. And I'll substitute in the values. So, a is the coefficient of x squared, b is the coefficient of x, and c is a constant coefficient. So, a is three, b is four, c is negative one. And so I'll substitute in those values into my quadratic formula. So, x equals negative b, b is four, plus or minus squared, b squared, b is four, minus four, a is three, c is negative one. The whole thing over two times a, that's my coefficient of x squared. So, there's my quadratic formula and at this point I can do some simple arithmetic. So, that's negative four doesn't change, plus or minus four, squared is sixteen, minus four times three times negative one, that's negative twelve, minus a negative twelve, that's plus twelve. Two times three is six. And my quadratic formula, negative four plus or minus square root is kind of like a parentheses symbol, sixteen plus twelve has to be done first, square root of twenty-eight. And it's helpful to remember that that plus or minus corresponds to two different solutions. So, I really have x negative four plus twenty-eight root or negative four minus square root of twenty-eight all over six. Well, have another example, x squared minus two x equals six. And in order to apply the quadratic formula I do need to get this into the form quadratic equal to zero. So, I need to subtract that six from both sides. So, I have x squared minus two x, I'll subtract six from both sides, on the right hand side I'll have zero and check it out, polynomial, degree two equal to zero. I can use the quadratic formula where I have a equals one coefficient of x, b equals the coefficient of x, negative two, c equal to our constant term is going to be negative six. So, I'll substitute these into my quadratic formula, I'll write it down and fill in the gaps. So, b is negative two, b negative two, a coefficient of x squared, that's going to be one, c constant term negative six, a coefficient of x squared is one. And I have my quadratic formula and I can fill in the values here. Negative, negative two, that's two, negative two squared, that's four, four times negative six, that's negative 24, I'm subtracting it two times one is two. So, four minus a negative four, that's 28, two stays the same, plus or minus squared to 28, and again it's helpful to remember that plus or minus corresponds to two solutions. So, I have two plus squared 28 over two and two minus squared to 28 over two.