can someone show me how to solve this sum, step by step

$
7x^2-25x-12=0
$

2. Originally Posted by 22upon7
can someone show me how to solve this sum, step by step

$
7x^2-25x-12=0
$
this can be factorized as (7x + 3)(x - 4) = 0

can you finish?

if you can't see the factorization so well, just use the quadratic formula

3. how did you get that answer?

4. Originally Posted by 22upon7
how did you get that answer?
well, in general if a quadratic can be factored, you can use the AC-method to factor it. do you know this method? we usually use it if the coefficient of x^2 is not 1, and is relatively big

here, 7 was not that big, so i decided to use trial and error. i know the only way to get 7 from multiplying integers is to have 7*1, so i set up the brackets

(7x )(x )

now, at the ends, i needed to put in numbers that i can multiply and get 12. 3 and 4 came to mind, so i tried them

(7x 3)(x 4)

now i needed to figure out the signs. we have -12, so the signs need to be different. moreover, i want the middle term to be -25x. but the middle term is calculated using the terms close together added to the terms far apart. that is, the middle term comes from some operation on 3x (=3*x) and 28x (=7x*4). if i have +3x and -28x then i get -25x, which is what i wanted, so i put those signs in accordingly.

(7x + 3)(x - 4)

of course, this is by trial and error, things won't always work out that smoothly, and you may have to switch numbers around and change numbers altogether. but since we are dealing with small numbers, it is not too much trouble

5. Originally Posted by 22upon7
can someone show me how to solve this sum, step by step

$
7x^2-25x-12=0
$
i was going to post that link, but it only has the case where the coefficient of $x^2$ is 1. i couldn't find the more general case (at least not the factoring method). maybe you should write one