# How do you solve the equation 2x^2+5x-3=0 by using the quadratic formula?

Dec 26, 2016

The two solutions are $x = 0.5$ and $x = - 3$

#### Explanation:

Since this question is given in standard form, meaning that it follows the form: $a {x}^{2} + b x + c = 0$, 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 $2 {x}^{2}$ for this question.$b$ is the number that has the $x$ variable associated with it and it would be $5 x$, and $c$ is a number by itself and in this case it is -3.

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

$x = \frac{- \left(5\right) \pm \sqrt{{\left(5\right)}^{2} - 4 \left(2\right) \left(- 3\right)}}{2 \left(2\right)}$

$x = \frac{- 5 \pm \sqrt{25 + 24}}{4}$

$x = \frac{- 5 \pm \sqrt{49}}{4}$

For these type of problems, you will obtain two solutions because of the $\pm$ part. So what you want to do is add -5 and $\sqrt{49}$ together and divide that by 4:

$x = \frac{- 5 + \sqrt{49}}{4}$
$x = \frac{2}{4} = 0.5$

Now, we subtract $\sqrt{49}$ from -5 and divide by 4:

$x = \frac{- 5 - \sqrt{49}}{4}$
$x = - \frac{12}{4} = - 3$

Therefore, the two possible solutions are:
$x = 0.5$ and $x = - 3$