Question

How do you use the quadratic formula to solve `2x^2-3x+2=0`?

Answer 1

No real solutions. Imaginary solutions: `x=(3+-sqrt(7)i)/4`

Explanation:

The quadratic formula allows us to solve a quadratic equation in the form `ax^2+bx+c` for `x` using the following:

`x=(-b+-sqrt(b^4-4ac))/(2a)`. So, generally, there will be two solutions.

For `2x^2-3x+2, a=2, b=-3, c=2`, and so

`x=(3+-sqrt((-3)^2-(4)(2)(2)))/(2*2)=(3+-sqrt(9-16))/4=(3+-sqrt(-7))/4`

Due to the negative root, there will be no real solutions, but the imaginary solutions would be

`x=(3+-sqrt(7)i)/4`