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x^2 - 3x - 4 = 0
factor it.

We first need to set it up in the form \(ax^2+bx+c=0\)

Ok I got that. I have x^2-3x-4=0 but I am not sure where to go from there.

Well that would be -4 and 1.

correct.

Correct

so, in
(x+b)(x-b)=0
form it would be?

So basically then we replace -3x with -4x +1x

So our equation would look as follows
\(x^2-4x+1x-4\)

Then we group the first 2 terms and teh last 2 terms
\( (x^2-4x)+(x-4)=0

\( (x^2-4x)+(x-4)=0\)

Then we factor out from each bracket
x(x-4)+1(x-4)=0

You made a mistake I believe.. @swissgirl

At which point?

x^2 - 4x + x -4
can just be simplified to
x^2 -3x - 4
.....

@Katelynmorris which method are you familiar with?

I am lost after the x^2-4x+1x-4=0..Solving by factoring

Ok so after the (x-4)(x+1)=0 that gives me x how?

(x-4)(x+1)=0
you put
x-4=0
and x+1=0
to get your solutions .

3x2−3x−4=0
swiss, i would use the quadratic formula. instead of trying to factor it.

You can use whtvr it makes no diff as long as you get your answer

Your method is much more complicated than it needs to be, just make it simple..

The only reason why I was showing him that method since that is the method taught in schools

Plus your method wont work for complicated equations trust me kid

Ok so x= 4, -1?

Yes, those are your solutions.

Thanks.

LAWL XD

OMG THEY ACTUALLY EXCIST o-o

stalking profile now

XD