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

Apr 17, 2016

$x = - 14 , 0$

#### Explanation:

$x = \frac{- \left(b\right) \pm \sqrt{{\left(b\right)}^{2} - 4 a c}}{2 a}$

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

$x = \frac{- \left(14\right) \pm \sqrt{{\left(14\right)}^{2} - 4 \left(1\right) \left(0\right)}}{2 \left(1\right)}$

Evaluate:
$x = \frac{- 14 \pm \sqrt{196 - 0}}{2}$
$x = \frac{- 14 \pm \sqrt{196}}{2}$
$x = \frac{- 14 \pm 14}{2}$

Solve to find the two roots of x
$x = \frac{- 14 + 14}{2} = \frac{0}{2} = 0$
$x = \frac{- 14 - 14}{2} = \frac{- 28}{2} = - 14$

$x = - 14 , 0$

Apr 17, 2016

#### 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:
${x}^{2} + 14 x = 0$
$x \left(x + 14\right) = 0$

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

The second possibility we may finish solving it:
$x + 14 = 0 \implies$ 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:
$- 14 \cdot \left(- 14 + 14\right) = - 14 \cdot \left(0\right) = 0$

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