Question

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

Answer 1

Rearrange to form: `3x^2+6x-10=0`
Substitute in quadratic formula: `x=(-(6)+-sqrt((6)^2-4(3)(-10)))/(2(3))`
Simplify to get: `x=(-3+sqrt(39))/(3) ~~1.08167` or `x=(-3-sqrt(39))/(3)~~-3.08167`

Explanation:

First of all you need to rearrange the equation into the form `ax^2+bx+c=0`

So first expand the brackets on the right hand side: `-3x^2=6x-10`

Then you can either take the `-3x^2` over or the `6x-10` over.

Let's just take the `-3x^2` over to give: `3x^2+6x-10=0`

Now we can see what `a` `b` and `c` are:
`a=3`
`b=6`
`c=-10`

Then we substitute these into the quadratic formula:
`x=(-b+-sqrt(b^2-4ac))/(2a)`
Quadratics usually have 2 answers which is why there is a `+-` as for one answer you use `+` and the other `-`.

So now substituting in: `x=(-(6)+-sqrt((6)^2-4(3)(-10)))/(2(3))`

This simplifies to: `x=(-6+-2sqrt(39))/(6)`

And dividing all terms by 2 to: `x=(-3+-sqrt(39))/(3)` which is the most simple form.

So `x=(-3+sqrt(39))/(3) ~~1.08167` or `x=(-3-sqrt(39))/(3)~~-3.08167`