Question

What are the zeros of the quadratic function `f(x)=6x^2+12x-7`?

Answer 1

`x = -1+-1/6 sqrt(78)`

Explanation:

`f(x) = 6x^2+12x-7`

is of the form `ax^2+bx+c` with `a=6`, `b=12` and `c=-7`

We can find the zeros of `f(x)` using the quadratic formula:

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

`color(white)(x) = (-12+-sqrt(12^2-4(6)(-7)))/(2(6))`

`color(white)(x) = (-12+-sqrt(144+168))/12`

`color(white)(x) = (-12+-sqrt(312))/12`

`color(white)(x) = (-12+-sqrt(4*78))/12`

`color(white)(x) = (-12+-2sqrt(78))/12`

`color(white)(x) = -1+-1/6 sqrt(78)`