In this example we're going to solve a quadratic equation that has complex number solutions using the quadratic formula. So here's the quadratic equation that we want to solve. 6x squared minus 2x equals minus 3. And so to apply the quadratic formula to solving this equation first we need to set the equation to zero.
So I'm going to do that by adding 3 to both sides of the equation. And that will make the right hand side 0. So we get 6x squared minus 2x plus 3 equals 0. And now to apply the quadratic formula I need to identify the coefficients in the equation. So the leading coefficient A is 6. The second coefficient B is 0. negative 2 and then the constant coefficient c is 3 and we're just going to substitute those into the quadratic formula as shown here and simplify to get the solutions to the equation.
So according to the quadratic formula the solutions to this equation are going to be x is minus b so that's minus a minus 2. Plus or minus, so again that's how we get two possible solutions to the equation. So plus or minus the square root of b squared, so that's minus 2 squared minus 4 times a is 6, and then c is 3, and then all of this divided by 2 times 6. All right so there are the two solutions to the equation. But of course we want to simplify this to get a better idea of what these two solutions actually are. So minus a minus 2 is plus 2. And then plus or minus the square root of negative 2 squared is 4. And then minus 4 times 6 is negative 24. And then negative 24 times positive 3 is negative 72. and then all of this divided by 12. So I end up here with 2 plus or minus the square root of negative 68. 4 minus 72 is minus 68, and then all of that divided by 12. And so here we can see now that the solutions, the two solutions to our equation are going to be complex numbers because In applying the quadratic formula to find solutions, we come up with this square root of a negative quantity.
And of course, that's going to be imaginary. And so that means that then that the solutions to our equation are going to be complex numbers. All right, now let's go ahead and simplify these two solutions as far as we can by hand. So let me show you how that process would go. All right, so first...
I'm going to take the minus 68 and I'm going to rewrite that as a product, but a very simple product here. I'm going to write that as minus 1 times positive 68. And the reason I'm doing that is just to show now that the solutions, again, to our equation are going to be complex, right? Because we can extract the square root of minus 1. We know that the square root of minus 1 is...
the imaginary number i. So we have 2 plus or minus the square root of, I'm sorry, 2 plus or minus i times the square root of 68 over 12. Now it's pretty good that we got that far, but actually we can simplify these two solutions a little bit further by hand also because we can simplify the square root of 68 in sort of the same fashion that we simplified the square root of minus 68. I'm going to take that 68 and write that as a product. And I want to write that as a product of a perfect square times some quantity. So if you think about 68 a little bit, you can see that that can be written as 4 times 17. So 68 is 4 times 17. And the reason it's beneficial to write 68 is 4 times 17 is 4 is a perfect square. So you can extract that square root of 4 just like we extracted the square root of minus 1 a moment ago.
And so the square root of 4, of course, is 2. So we get a further simplification here of our two solutions. We have 2 plus or minus 2 times i times the square root of. 17 because we can extract that square root of 4 as 2 in front of the radical sign. All of this divided by 12. Now you won't be able to simplify square root of 17 at all. 17 is a prime number.
But we can still continue simplifying these two solutions to our quadratic equation because we can factor 2 from both of the terms in the numerator. So we can write the 2 plus or minus 2i times the square root of 17 as 2 times 1 plus or minus i times the square root of 17 over 12. So I'm going to factor a 2 out of both of these terms in the numerator. And the reason that's beneficial is now I can divide 2 into 12, so I can reduce 2 over 12, and so I end up here with 1 plus or minus i times the square root of 17 over 6. 2 goes into 12 six times, of course. So there are the two solutions, fully simplified, as far as I can by hand at any rate, to my original quadratic equation.
One of my solutions is 1 plus i times the square root of 17 divided by 6, that complex number. And then the second solution would be 1 minus i times the square root of 17 divided by 6. So that quadratic equation that we solved had two solutions. Both of them are complex numbers.