How do you solve `5m^2 - 11m - 3 = 0` using the quadratic formula?

So, when we have a quadratic equation:

`ax^2 + bx + c = 0`

One of the ways we can solve for `x` is by using the quadratic formula:

![https://www.blendspace.com/lessons/fk5Tc7PpFlsQWg/the-quadratic-formula]()

But wait... for your problem, there's no `x`! Taking a look at the equation, the first thing we need to do is realize that `x` is just a placeholder for any variable (a number that we don't know... yet). In your question, the variable is shown as `m`. It's the same thing; we could call it `x` or `m` or `omega` or whatever we like.

Now, we need to figure out the values of `a`, `b`, and `c`. `a` is the number before the `x^2` variable (or, in your case, `m^2`). `b` is the number before `m`, and c is the number not associated with the variable. So...

`a = 5`

`b = -11` (don't forget that negative sign!)

`c = -3` (again, the negative sign is important)

Now, we can take those three numbers and plug them into the quadratic equation:

`x = (-(-11) (+-) sqrt((-11^2)- (4)(5)(-3))/((2)(5)))`

See that `+-` in there? That means we're going to get two answers.

Either

`x = (-(-11) (+) sqrt((-11^2)- (4)(5)(-3))/((2)(5)))`

OR

`x = (-(-11) (-) sqrt((-11^2)- (4)(5)(-3))/((2)(5)))`

`x` can either be 2.445 or -0.2453 (leaving off quite a few decimal places)

Let's check to make sure that's right, by plugging in our values for `x`:

Does `(5)(2.445^2) + (-11)(2.445) + (-3) = 0`?

Does `(5)(0.2453^2) + (-11)(0.2453) + (-3) = 0`?