8x^2-2x-3=0, please tell me the way how you did it.

Question

anonymous · Answer

believe it or not this one factors, but you can always use 
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ 
if it is not clear how to factor it

anonymous · Answer

Factor to get:
$$(4x-3)(2x + 1)\ =\ 0$$

anonymous · Answer

$$4x -3 = 0\ \ OR\ \ 2x +1 =0$$
$$x = \frac{3}{4}\ \ or \ x = -\frac{1}{2} $$

anonymous · Answer

This one could be done by grouping i believe.  8 times -3 is -24, and the middle number is -2, so you want to think of two numbers that multiply to -24 and add to -2.
These numbers are -6 and 4.  So now we change that -2x to (-6x+4x) and we get:
$$8x^2-2x-3 = 8x^2+4x-6x-3 = 4x(2x+1)-3(2x+1) $$
$$= (4x-3)(2x+1)$$

anonymous · Answer

i love "factor by grouping" but i can never to it!

mathteacher1729 · Answer

You could try factoring: 


Here is my thought process: 

(8x + a ) ( x + b ) = 8x^2-2x-3
or 
(2x + a)( 4x + b) = 8x^2-2x-3

Now the choices of a and b can be anything in the set 
{ +1 , -1 , +3 , -3}

anonymous · Answer

i learned how to do it on this site lol, someone else said what a posted and i was like "OH SO THATS HOW YOU DO IT!!" lolol

anonymous · Answer

hey joemath you are completely right but how did u get 4x,-6x?

anonymous · Answer

i got it men thank u.