Question

How do you use the quadratic formula to solve `3x^2- 8x -15 =0`?

Answer 1

`x=3.94" and "-1.27` to 2 decimal places (approx. solution)

Explanation:

This is really worth trying to remember. I did it many years ago by making a point of writing it out for every question I attempted. Repetition is the key.

The standard form is:

`y=ax^2+bx+c" "` where `" "x=(-b+-sqrt(b^2-4ac))/(2a)`
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Given: `3x^2-8x-15=0`

Set `a=3"; "b=-8"; "c=-15`

So by substitution we have:

`x=(+8+-sqrt((-8)^2-4(3)(-15)))/(2(3))`

`x=(8+-sqrt(64+180))/6`

`x=4/3+-sqrt(244)/6`

`x=4/3+-sqrt(2^2xx61)/6`

`x=4/3+-sqrt(61)/3" "` This is an exact solution

`x=3.94" and "-1.27` to 2 decimal places (approx. solution)