Question

How do you solve `2x^2 - 3x + 4 = 0` using the quadratic formula?

Answer 1

`x = 3 pm sqrt(7)/(2sqrt(2))`

Explanation:

`f(x) = (xsqrt(2) -3/sqrt(2))^2 + 4 - 9/4 = 0`
if `xsqrt(2) -3 /sqrt(2) = pm sqrt(7)/4`
or `x = 3 pm sqrt(7)/(2sqrt(2))`

Answer 2

`x = 3/4+-sqrt(23)/4i`

Explanation:

Given:

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

Note that this is of the form:

`ax^2+bx+c = 0`

with `a=2`, `b=-3` and `c=4`

The discriminant `Delta` of this quadratic is given by the formula:

`Delta = b^2-4ac = (-3)^2-4(2)(4) = 9-32 = -23`

Since `Delta < 0` we can tell that this quadratic has no Real zeros - only Complex ones.

We can still use the quadratic formula to find them:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-b+-sqrt(Delta))/(2a)`

`color(white)(x) = (3+-sqrt(-23))/4`

`color(white)(x) = (3+-sqrt(23)i)/4`

`color(white)(x) = 3/4+-sqrt(23)/4i`

Answer 3

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

The solutions to `ax^2+bx+c=0` are given by

`x = (-b +- sqrt(b^2-4ac))/(2a)`.

In this case, we get

`x = (-(-3)+-sqrt((-3)^3-4(2)(4)))/(2(2))`

`= (3 +- sqrt(9-32))/4`

`= (3 +- sqrt (-23))/4`

The solutions are imaginary

`3/4 +- sqrt23/4 i`