# Thread: Need help solving for:

1. $\displaystyle 6x^2-5x-6$
Factors of 6 are:
3,2
1,6
Factorizing then gives $\displaystyle (2x-3)(3x+2)$
If $\displaystyle (2x-3)(3x+2)=0$,
$\displaystyle x=3/2$ or $\displaystyle x=-2/3$
Which part makes it confusing? Is it the $\displaystyle 6x^2$?

2. Originally Posted by Vuong
You could use quadratic to factor?
I believe I just did.

3. Originally Posted by Quacky
$\displaystyle 6x^2-5x-6$
Factors of 6 are:
3,2
1,6
Factorizing then gives $\displaystyle (2x-3)(3x+2)$
If $\displaystyle (2x-3)(3x+2)=0$,
$\displaystyle x=3/2$ or $\displaystyle x=-2/3$
Which part makes it confusing? Is it the $\displaystyle 6x^2$?
Yes, for that I just kept trying (6x + ) (x )

4. Sometimes, that will give you a solution, but you must consider the other possible ways of getting $\displaystyle 6x^2$ as well.

This website is rather babyish, but the explanations are clear, and it makes factorizing a lot easier.

5. How do I find the x intercepts of a problem like y=x^2-6x+25?

6. Originally Posted by Quacky
y=-4x^2-9x-2

Ok, the line will cut the y axis where x=0.
Substitute x=0 into the equation, and you get y=-2. So point of interception is at:
(0,-2)
Now, the line will cut the x axis where y=0. To find this, we can try to factorise:

-4x^2-9x-2=0
4x^2+9x+2=0
(4x+1)(x+2)=0
x=-2 and x=-1/4
x intercepts are
$\displaystyle (-2,0)$ and $\displaystyle (-1/4,0)$

Parabolas are symmetrical, so if you take the x intercepts, the line of symmetry will be halfway between them.
$\displaystyle -2.25/2 = -1.125$ (wow, nasty numbers)
x=-1.125 is the line of symmetry.

A question about the vertex: Have you covered differentiation?
Also, how would I graph this as a parabola? If both x are on the negative side

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