Question

How do you solve `x^2 – 3x = 4x – 1` using the quadratic formula?

Answer 1

First, we have to move everything to the left so it resembles this form: `ax^2+bx+c=0`, also known as the standard form.

For this case, all we have to do is move everything to 1 side (it doesn't matter which side, although I prefer to move it all to the side with `x^2` already in it)

`x^2-3x=4x-1`
`x^2-7x+1=0`

Now you will need to know the quadratic formula to solve this problem: `x=(-b+-sqrt(b^2-4ac))/(2a)`

But what are `a`, `b`, and `c` you ask? Well that's why we rearranged the original equation. `a` is the number next to `x^2`, `b` is the number next to `x`, and `c` doesn't have an `x` next to it. Doing this, we find `a=1`, `b=-7`, `c=1`. All we have to do is put into the quadratic formula and simplify:

`x=(-(-7)+-sqrt((-7)^2-4(1)(1)))/(2(1))`
`x=(7+-sqrt(49-4))/(2)`
`x=(7+-sqrt(45))/(2)`

So this gives us two answers:
`x=(7+sqrt(45))/(2) or (7-sqrt(45))/(2)`
Which are the correct answers.