Question

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

Answer 1

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: `ax^(2) + bx + 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 `2x^(2)` for this question.`b` is the number that has the `x` variable associated with it and it would be `5x`, 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 = (- (5) +- sqrt((5)^(2) - 4(2)(-3)))/(2(2))`

`x = (-5 +-sqrt(25+24))/4`

`x = (-5 +- sqrt(49))/4`

For these type of problems, you will obtain two solutions because of the `+-` part. So what you want to do is add -5 and `sqrt(49)` together and divide that by 4:

`x = (-5+sqrt(49))/4`
`x = 2/4= 0.5`

Now, we subtract `sqrt(49)` from -5 and divide by 4:

`x = (-5-sqrt(49))/4`
`x = -12/4 = -3`

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