Question

Factor: 3n^2+12n-96=0

Answer 1

use the quadratic formula
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
since the standard quadratic form is
ax^2=bx+c=0
the a will be replaced by 3, 12 by b, and c by -96
a=3
b=12
c=-96

now plug and chug :3

Answer 2

So, instead of it being the standard quadratic form, I use that one? Ok, I'll see if I can make this work. I used to do these all the time, but it's been a while so I'm a little rusty. If I need your help again I'll tag you. Thank you. c:

Answer 3

Sure :)

Answer 4

The -4ac in the formula is confusing me a bit.

Answer 5

Would it be 4(3)(-96)?

Answer 6

\[x=\frac{ -(12)\pm \sqrt{12^2-4(3)(-96)} }{ 2(3) }\]
4(3)(-96)=12*-96=-1152
\[x=\frac{ -12\pm \sqrt{144-(-1152)} }{ 6}\]

Answer 7

\[x=\frac{ -12\pm \sqrt{144+1152} }{ 6 }\]

Answer 8

makes sense? :)

Answer 9

Yeah, I'm taking these notes. Thank you. c:

Answer 10

No problem

Answer 11

are you ok with the next steps?

Answer 12

I should be c:

Answer 13

Good c:

Answer 14

right? & then find the square root & divide both 12 & the square root by 6?