Question

Using the quadratic formula to solve 5x=6x^2-3, what are the values of x?

Answer 1

`x=(5 \pm √(97))/(12)`

Explanation:

First step is to arrange so it is in the form
`ax^2+bx+c=0`.

We have
`5x=6x^2-3`.

Add we really need to do is subtract 5x on both sides:

`0=6x^2-5x-3`.

Now let's compare
`6x^2-5x-3` to
`ax^2+bx+c`.

We have
`a=6,b=-5,c=-3`.

The quadratic formula is
`x=(-b \pm √(b^2-4ac))/(2a)`.

I like to break this into parts:

Part 1: Find
`-b`.

Part 2: Find
`b^2-4ac`.

Part 3: Find
`2a`.

Answering the parts:

Part 1:
`-b=5` since
`b=-5`.

Part 2:
`b^2-4ac=(-5)^2-4(6)(-3)=25-24(-3)=25+72=97`.

Part 3:
`2a=2(6)=12`.

Now our formula in terms of my parts looks like this:

`x=\frac{\text{Part 1} \pm √(Part 2)}{Part 3}`

Our formula with my parts evaluated looks like this:

`x=(5 \pm √(97))/(12)`.