Question

How do you solve `3x^2-10x-8=0`?

Answer 1

`x = -2/3 or 4`

Explanation:

You can just eyeball it most of the time, for instance I can quickly see that this will factorise to

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

In order for the left hand side to be equal to zero then one of the brackets must be equal to zero, so we have

`3x+2 = 0 or x - 4 = 0`

`implies x = -2/3 or 4`

This isn't particularly handy if it's a really nasty equation or if you just can't see the solution so in general you should use the quadratic formula. For quadratic of the form:

`ax^2+bx+c = 0`

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

So in this case:

`x = (10+-sqrt(10^2-4(3)(-8)))/(2(3)) = (10+-sqrt(196))/6 = (10+-14)/6`

`therefore x = -4/6 = -2/3 or x = 24/6 = 4`

Answer 2

`x = -2/3, 4`

Explanation:

since you have a number greater than 1 before the `x^2` you will need to multiply that by the `-8` at the end to get true divisible numbers. 3 x -8 = -24, ergo you need to find two numbers that multiply to get -24 and add up to reach -10 (your middle term). in this case, it would be 2 and -12.

align the equation to make further factoring easier:

`3x^2 - 10x - 8 = 0`

`3x^2 - 12x + 2x - 8 = 0` note that `(3x^2 + 2x - 12x - 8 = 0)` isn't as user friendly

factor the `(3x^2 - 12x + 2x - 8 = 0)` into two sets of brakcets:

`(3x (x - 4) + 2(x - 4) = 0)` and group them together:

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

now you need to determine what values for `x` will make this equation true.
`3x + 2 = 0`

`3x = -2`

`x = -2/3`

AND

`x - 4 = 0`

`x = 4`