Question

How do you solve `2x^2+6x+33=0`?

Answer 1

`x = -3/2+-sqrt(57)/2i`

Explanation:

`2x^2+6x+33 = 0`

This is in the form:

`ax^2+bx+c = 0`

with `a = 2`, `b = 6` and `c = 33`.

The discriminant `Delta` is given by the formula:

`Delta = b^2-4ac = color(blue)(6)^2-4(color(blue)(2))(color(blue)(33)) = 36 - 264 = -228 = -2^2*57`

Since `Delta < 0` this quadratic has a Complex conjugate pair of roots, which we can find using the quadratic formula:

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

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

`color(white)(x) = (-6+-sqrt(-2^2*57))/(2*2)`

`color(white)(x) = (-6+-2sqrt(57)i)/4`

`color(white)(x) = -3/2+-sqrt(57)/2i`