 One of the interesting things about mathematics is that mathematics spawns more mathematics, and in the case of the quadratic formula, this is centered around something called the discriminant. So let's review that, our quadratic formula, if I have a quadratic equation, I can find the solutions using the quadratic formula negative b plus or minus squared b squared minus 4ac, the whole thing over to a. Now notice that in this particular equation, this formula, there's two potential problem areas. First of all, because it's a quotient, we're dividing by 2a, we have to make sure that 2a can't be equal to zero. Now this is only going to occur if a is equal to zero, and if we actually have a quadratic, if we actually have a square term, that means a is not zero. Which means that we don't really have to worry about this, 2a is not going to be zero. But in another way, if you ever run into a problem where 2a is zero, you're not dealing with the quadratic, and the quadratic formula doesn't really apply. The bigger problem is we're taking the square root of this quantity, b squared minus 4ac, and because we're taking a square root, we need this number to be a positive number. You cannot take the square root of a negative number, or can you? Well, we'll see about that. But for right now, we need that square root, we need that radicand b squared minus 4ac to be positive. And sometimes it's not going to be. And because of that, this expression b squared minus 4ac is called the discriminant of the quadratic formula. Now if you continue to study mathematics, you'll look at other discriminants, so we do want to be specific here, this is the discriminant for the quadratic formula. And whether this is positive or negative or zero, it's going to tell us something about the solutions to our quadratic equation. So there's three important cases. So if b squared minus 4ac is positive, then this is a real number, and we actually get two real solutions, x equals this, x equals that. If it's zero, then the square root is also going to be zero, and I'll have two solutions again, negative v plus zero over 2a, negative v minus zero over 2a, except these are the same number. When I add or subtract zero, I get the same thing. And so while there are two solutions, they happen to be the same thing. And so we say there's one distinct real solution. And then finally, if b squared minus 4ac is negative, this isn't a real number, and so I have no real solutions. For example, let's say I have the quadratic x squared minus 5x plus 12. Well, I could compute the discriminant only, but we might as well find the actual solution. There's no harm in computing it, and it's good practice using the quadratic formula, which you want to memorize in any case. So we have our quadratic formula, where again a is the coefficient of x squared, that's 1, b is the coefficient of x, that's negative 5, c is the constant that's plus 12, so I can substitute those into the quadratic formula, b negative 5, negative 5, a1, c12, a1, and I get this nice little expression, which I can evaluate using the operations of integer arithmetic, that's 5, plus or minus square root of 25 minus 48, and after all the dust settles, 5 plus or minus square root of, uh oh, negative 23 is our square root. And since square root of negative 23 is not a real number, there are no real solutions. The solutions involve this, what is this square root of negative? Well, I'll have to take a look at that a little bit later on, but it's not a real number. Well, this also translates into geometric problems, find the x intercepts of the graph, and let's take a look at that. So again, if I want to find the x intercepts of the graph, I am looking at points on the x axis, and if I'm on the x axis, the y coordinate is zero. So we're looking for the x values when y is equal to zero. So if y is zero, there it is. I want to know the x values that will make this equation true. Well, that's a quadratic equation. So in order for these points to correspond to real points, it's necessary that those x coordinates actually be real numbers. So let's go ahead and substitute that in. So there's my quadratic equation, y zero equals 12 minus 5x minus 3x squared. A is the coefficient of x squared. Notice that I can use a negative coefficient. That's not a problem. B, coefficient of x, negative 5, c, constant, that's going to be 12, and so I can fill the quadratic formula in x equals negative b plus or minus squared, b squared minus 4ac, all over 2 times a. And so after the dust settles here, this is going to be 5 plus or minus squared 169 over negative 6, and squared 169 is actually a real number. Square root only makes sense if I'm taking the square root of a positive number. 169 is definitely positive. In fact, squared 169 is 13. So this equation has two real solutions. The graph crosses the x axis twice, once at x 5 plus 13 over negative 6, and the other at 5 minus 13 over negative 6.