Question

How do you solve `x^2+14x=0` using the quadratic formula?

Answer 1

`x = -14, 0`

Explanation:

Quadratic Formula
`x = (-(b) pm sqrt{(b)^2 - 4ac})/{2a}`

Substitute:
a term = 1
b term = 14
c term = 0

`x = (-(14) pm sqrt{(14)^2 - 4(1)(0)})/{2(1)}`

Evaluate:
`x = (-14 pm sqrt{196 - 0})/2`
`x = (-14 pm sqrt{196})/2`
`x = (-14 pm 14)/2`

Solve to find the two roots of x
`x = (-14 + 14)/2 = 0/2 = 0`
`x = (-14 - 14)/2 = (-28)/2 = -14`

`x = -14, 0`

Answer 2

Check my full answer, please.

Explanation:

Not disagreeing with Dèv, but there is another(and even easier) way to solve it:

In a quadratic function, whenever `c=0` its possible to solve it this way. First, you put `x` in evidence and then you find 2 possibilities:
`x^2+14x = 0`
`x(x+14)=0`

So, the 2 possibilities: `x = 0` or `x+14 = 0`. The first possibility is already solved, because if `x=0`:
`0*(0+14) = 0*(14) = 0`. As we can see, its correct.

The second possibility we may finish solving it:
`x+14 = 0 =>` The `14` is positive and goes to the other side of the equality as `-14`, but `0-14 = -14`, so `x = -14`.

Now, we do solve it the same way we did it before:
`-14*(-14+14) = -14*(0) = 0`

So, the roots of a quadratic function with `c = 0` are:
`ax+b = 0` and `x = 0`