Hey everyone. You just finished learning 3 different methods of solving quadratic equations, and you might be wondering how there's anything left and why we have yet another method to learn. But the quadratic formula that we're going to talk about now is really great because it's going to work for any quadratic equation. So even if you were to forget every single other method, as long as you remember the quadratic formula, you're going to be able to solve any quadratic equation that gets thrown at you. So let's go ahead and jump in. The quadratic formula is based on the standard form of a quadratic equation ax2+bx+c, and we're going to use a, b, and c in order to compute our solutions. So the quadratic formula is −b±b2−4×a×c2×a. Now that formula might look a little bit complicated right now and, unfortunately, it is something that you're going to have to memorize. But a way that I was able to memorize the quadratic formula was by using the quadratic formula song. Now, I'm not going to sing it for you right now, but it's going to be something that you should search up on your own to commit this formula to memory.
So when do we want to use the quadratic formula? Well, like I said, you can use the quadratic formula whenever you want, and some clues that you might want to use it are that you can't easily factor or you're just otherwise unsure what method to use. Let's go ahead and take a look at an example here. So I have x2+2x−3 is equal to 0. Let's go ahead and take a look at our first step, which is to write our equation in standard form. Now it looks like my equation is already in standard form, all of my terms are on the same side in descending order of power, so I can go ahead and move on to step 2. Now step 2 is simply to just plug everything into my quadratic formula, so let's go ahead and do that here. I'm first going to label a, b, and c in my equation so that it's easy for me to just take them and plug them in. So in this case, I have an invisible one in front of that x2 and that is my a, b is this positive 2, and then c is negative 3. Make sure that you're paying attention to the signs here. So plugging this in, I get −2±22−4×1×−32×1. So now that we've plugged everything in, we've completed step 2, and all that we have left to do from here is algebra. We're just going to compute and simplify our solutions now.
So let's start with what's in our radical here and just simplify that. So this 22 is going to become 4 minus 4 times 1 is just 4, so, really, that's just 4 times negative 3. I know that 4 times negative 3 is negative 12, and 4 minus negative 12, this becomes a plus, so this is really just 16. So all of that under the radical is just 16. Let's go ahead and rewrite our formula with a little more simplifying. So this is −2±162, because that's what we just found under the radical was, all divided by 2 times 1, which is just 2. Okay. So from here, I can do a couple more things. Now I know that the square root of 16 is just 4, so this becomes negative 2 plus or minus 4 all divided by 2. Okay. Now is when we're going to want to split our solution into the plus and the minus. So let's go ahead and do that. So this splits into negative 2+4 over 2 and negative 2 minus 4 over 2. And now I can just simplify those solutions down. So negative 2+4 is going to give me positive 2 over 2, which is just equal to 1. And then negative 2 minus 4 is going to give me negative 6 divided by 2, which is going to give me negative 3. And I'm done. These are my solutions. So for this quadratic equation, my solutions are x equals 1 and x equals negative 3 and I'm completely done there. I know that the quadratic formula can look a little intimidating at first but it really just comes down to the algebra once you plug everything in. Let's go ahead and take a look at one more example.
So here I have x2−5x is equal to negative one. Let's go ahead and start back at step 1, which is to write our equation in standard form. Now here, it looks like I need to go ahead and move my negative one over. So I can do that by simply adding 1 to both sides. And so this will become x2−5x+1 is equal to 0, because that one canceled on that right side. So step 1 is good. That's in standard form now. We can move on to step 2 and just plug everything in. So again, I'm going to go ahead and label a, b, and c here. I again have an invisible one for that a, b is negative 5, and c is positive 1. So let's go ahead and complete step 2 and plug everything in. So I have negative negative 5, remember to include that sign, plus or minus the square root of negative 5 squared minus 4 times a, which is 1, times c, which is also 1. Okay. And that's all divided by 2 times a, which again is 1. Okay. So step 2 is done. Now we're just left to simplify. Just do all of that algebra. So let's first start with everything under that radical. So negative 5 squared is going to give me 25 minus 4 times 1 times 1 is just 4, so this is just 25 minus 4, which gives me 21. So everything under the radical just simplifies to 21. Let's rewrite our quadratic formula. So negative negative 5 is going to be positive 5, plus or minus the square root of 21, that's what we just found, divided by 2 times 1, which is just 2. Okay, so looking at this I'm actually completely done and I can't simplify this anymore. Even if I split it, nothing would simplify because I just have that square root of 21 which can't go anywhere from there. So my solutions are just 5 plus or minus the square root of 21 over 2. Now I can of course split these into their plus and minus just to show both answers separately and I would just write 5 +root21 over 2 and 5 minus root 21 over 2. Now, as you saw with some of our other methods, you're going to have different types of answers here. So you might have whole numbers, you might also have some combination of fractions and radicals, and either is fine. That's all there is to the quadratic formula let me know if you have any questions.
