How do you find the zeros of `g(x) = 4x^2-2x-3` using the quadratic formula?

`g(x) = 4x^2-2x-3`

is in the standard form `ax^2+bx+c`, with `a=4`, `b=-2` and `c=-3`

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

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

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

`color(white)(x) = (2+-sqrt(4+48))/8`

`color(white)(x) = (2+-sqrt(52))/8`

`color(white)(x) = (2+-sqrt(2^2*13))/8`

`color(white)(x) = (2+-2sqrt(13))/8`

`color(white)(x) = 1/4+-sqrt(13)/4`

`color(white)()`
Footnotes

Given any quadratic equation in the form `ax^2+bx+c = 0`,

the quadratic formula:

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

is very useful, but do you know where it comes from, how it works or how to derive it yourself?

Here's one way...

Given `ax^2+bx+c = 0`, (with `a != 0`) we have:

`0 = ax^2+bx+c`

`color(white)(0) = a(x^2+2b/(2a)x + b^2/(4a^2) - b^2/(4a^2) + c/a)`

`color(white)(0) = a((x+b/(2a))^2-(b^2-4ac)/(4a^2))`

`color(white)(0) = a((x+b/(2a))^2-(sqrt(b^2-4ac)/(2a))^2)`

`color(white)(0) = a((x+b/(2a))-(sqrt(b^2-4ac)/(2a)))((x+b/(2a))+(sqrt(b^2-4ac)/(2a)))`

`color(white)(0) = a(x+(b-sqrt(b^2-4ac))/(2a))(x+(b+sqrt(b^2-4ac))/(2a))`

Hence:

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