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

Dec 26, 2016

The two solutions are $r = \frac{- 1 + \sqrt{113}}{4}$ and $r = \frac{- 1 - \sqrt{113}}{4}$

or $r = 2.4075$ and $r = - 2.9075$

#### Explanation:

Since this question is given in standard form, meaning that it follows the form: $a {x}^{2} + b x + c = 0$, or $a {r}^{2} + b r + c = 0$ in this case, we can use the quadratic formula to solve for x: I think it's worthwhile to mention that $a$ is the number that has the ${x}^{2}$ term associated with it. Thus, it would be $2 {r}^{2}$ for this question.$b$ is the number that has the $x$ variable associated with it and it would be $1 r$, and $c$ is a number by itself and in this case it is -14.

Now, we just plug our values into the equation like this:

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

$r = \frac{- 1 \pm \sqrt{1 + 112}}{4}$

$r = \frac{- 1 \pm \sqrt{113}}{4}$

For these type of problems, you will obtain two solutions because of the $\pm$ part. So what you want to do is add -1 to $\sqrt{113}$ together and divide that by 4:

$r = \frac{- 1 + \sqrt{113}}{4}$
$r = \frac{9.6301}{4} = 2.4075$

Now, we subtract $\sqrt{113}$ from -1 and divide by 4:

$r = \frac{- 1 - \sqrt{113}}{4}$
$r = - \frac{11.6301}{4} = - 2.9075$

Therefore, the two possible solutions are:
$r = 2.4075$ and $r = - 2.9075$