Question

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

Answer 1

The two possible solutions are
`x = 3`
`x = -0.50`

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 +- 7)/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 7 together and divide that by 4:

`x = (5+7)/4`
`x = 12/4 = 3`

Now, we subtract 7 from 5 and divide by 4:

`x = (5-7)/4`
`x = -2/4 = -0.50`

Next, plug each value of x into the equation separately to see if your values give you 0. This will let you know if you performed the calculations correctly or not

Let's try the first value of `x` and see if we obtain 0:

`2(3)^(2)-5(3)-3 = 0`

`18 - 15 - 3 =0`

`0= 0`

Thus, this value of x is correct since we got 0!

Now, let's see if the second value of `x` is correct:

`2(-0.50)^(2)-5(-0.50)-3 = 0`

`0.50 -2.5 - 3 = 0`

`0= 0`
That value of x is correct as well!

Therefore, the two possible solutions are:

`x = 3`
`x = -0.50`

Answer 2

`x=-1/2, 3`

Explanation:

Solve the quadratic equation `2x^2-5x-3=0` for `x` using the quadratic formula. The quadratic equation in standard form is `ax^2+bx+c`, where `a=2`, `b=-5`, and `c=-3`.

Quadratic Formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Plug the given values into the formula and solve.

`x=(-(-5)+-sqrt((-5)^2-4*2*-3))/(2*2)`

Simplify.

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

Simplify.

`x=(5+-sqrt49)/4`

`x=(5+-7)/4`

Solve for `x`.

There a two equations.

`x=12/4` and `x=-2/4`

Simplify.

`x=3` and `=-1/2`

`x=-1/2, 3`