Question

`6r^2+r-12=0`?

Answer 1

`r=4/3,-3/2`

Explanation:

We got: `6r^2+r-12=0`

Using the quadratic formula, the roots of a quadratic equation `(ax^2+bx+c=0)` are given by:

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

So here, we got `a=6,b=1,c=-12`, and so

`r=(-1+-sqrt(1-4xx6xx(-12)))/12`

`r=(-1+-sqrt(1+288))/12`

`r=(-1+-sqrt(289))/12`

`r=(-1+-17)/12`

`r=cancel16^4/cancel12^3,cancel(-18)^-3/cancel12^2`

`r=4/3,-3/2`

So, the roots of the equation `6r^2+r-12=0` are: `r=4/3` and `r=-3/2`.

Answer 2

`r = 4/3 , -3/2`

Explanation:

`6r^2 + r - 12 =0`

Use the splitting the middle term method.

`6r^2 + 9r -8r-12 =0`

`3r(2r +3) -4(2r+3) = 0`

`(3r-4)(2r+3) =0`

So , `3r -4=0`

`3r =4`

`r=4/3`

Or

`2r+3 =0`

`2r =-3`

`r=-3/2`

Answer 3

See a solution process below: `r = -3/2` and `r = 4/3`

Explanation:

If you are trying to solve for `r` you can use the quadratic equation to solve this problem:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(6)` for `color(red)(a)`

`color(blue)(1)` for `color(blue)(b)`

`color(green)(-12)` for `color(green)(c)` gives:

`r = (-color(blue)(1) +- sqrt(color(blue)(1)^2 - (4 * color(red)(6) * color(green)(-12))))/(2 * color(red)(6))`

`r = (-1 +- sqrt(1 - (-288)))/12`

`r = (-1 +- sqrt(1 + 288))/12`

`r = (-1 +- sqrt(289))/12`

`r = (-1 - 17)/12` and `r = (-1 + 17)/12`

`r = -18/12` and `r = 16/12`

`r = -3/2` and `r = 4/3`