Question

How do you solve `2x^2 - 3x + 1 = 0`?

Answer 1

`x = 1/2, 1`

Here's how I solved it:

Explanation:

You solve this by using the quadratic formula. This equation is written in standard form, or `ax^2 + bx + c = 0`. The quadratic formula is `x = (-b +- sqrt(b^2-4ac))/(2a)`.

So let's plug those numbers in and solve for `x`:
`x = (-(-3) +- sqrt((-3)^2 - 4(2)(1)))/(2(2))`

`x = (3 +- sqrt(9 - 8))/4`

`x = (3 +- sqrt(1))/4`

`x = (3 +- 1)/4`

So `x = (3+1)/4` and `(3-1)/4`.

`x = 4/4 -> x = 1`

`x = 2/4 -> x = 1/2`

`x = 1/2, 1`

Hope this helps!