 
Welcome back to speller tutorial services in today's video. We're going to use the quadratic formula to solve a quadratic equation that has
complex or Imaginary roots
Alright, let's begin by identifying the a b and C. That are necessary for the quadratic equation or the quadratic formula
Here the coefficient in front of the x squared term and that's going to give us a in this case. That's
5b is going to be the coefficient of the x term which is 2 and C is going to be the value of 3
the formula [for] the
Or the quadratic formula is negative B plus
or minus
B squared
minus 4ac and that whole quantity divided by the expression to a
Now let's make our substitutions be here has the value of 2V
squared is 2 times 2 which gives us 4
Minus 4 times a which is five times C. Which is three and
That's going to be divided by [two] times a and in this case a is
five
Now we'll begin by will continue by evaluating. What's underneath the radical here
so I have four minus well 4 times 5 is 20 and
20 times 3 is 60 and
Then the denominator here is 2 times 5 or just 10?
Will continue to evaluate what's underneath the radical
I have 4 minus 60 which gives me a negative 56 all
of this here 10
Will continue appear at the top?
we're going [to] take that that negative 56 and
[I'm] going to do the prime factorization. I know that 8 times 7 gives me 56
To get 8 that's four times two and to get two. That's two times two
So now I've breaking that 56 down into all of its prime
Factors and after remember to negative so I know that I'm going to also have a negative one here
So what I end up with is
2
plus or minus
2 Times 2 Times 2
Times 7 times negative 1 and
We're all of that is divided by 10, and I know with a square root. I'm looking for pairs
all right, let me rewrite that negative 2 plus or minus and
With the square root of looking for pairs here. I have a pair of two's so that means that two escapes
I also have a negative one which comes out as I and then the leftovers are 2 times 7
Or just 14, and then my denominator here is 10
I see that my negative [2] my two and
Attended my denominator all share a gcf of two so that means I can divide each one of those numbers by two
Well negative 2 divided by 2 gives me negative
12 divided by 2 gives me a positive one which I don't need to write because I'm multiplying and then in my denominator
10/2 gives me five so my complex root for this particular
quadratic
Equation is negative the quantity negative 1 plus or minus. I route 14 where all of that is divided by 5
All right, that sums it up for our
for this particular video
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