Question

How do you find the roots, real and imaginary, of `y= 12x^2 + 35x + 72` using the quadratic formula?

Answer 1

`x = -35/24+-sqrt(2231)/24i`

Explanation:

The equation:

`y = 12x^2+35x+72`

is a quadratic in standard form:

`y = ax^2+bx+c`

with `a=12`, `b=35` and `c=72`

It has zeros given by the quadratic formula:

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

`color(white)(x) = (-(color(blue)(35))+-sqrt((color(blue)(35))^2-4(color(blue)(12))(color(blue)(72))))/(2(color(blue)(12)))`

`color(white)(x) = (-35+-sqrt(1225-3456))/24`

`color(white)(x) = (-35+-sqrt(-2231))/24`

`color(white)(x) = -35/24+-sqrt(2231)/24i`

Note that `2231 = 23 * 97` has no square factors, so `sqrt(2231)` cannot be simplified.