Question

solve the equation:
70z^2-5z-15

Answer 1

First step factor out 5 from each term

Answer 2

5(14z^2 -z -3)

Answer 3

can factor but without an= something, can not solve

Answer 4

quadratic formula. (-b+/- sqrt(b^2-4ac))/2a

x= .5 or x=-.4285

Answer 5

sorry it equals 0

Answer 6

To clarify the formula... you have it set up as...

Az^2+Bz+C=0 where A B and C are all constants
70z^2+(-5)z+(-15)=0

Answer 7

then you take out the five and set up parenthesis
but then what

Answer 8

so it's set up...(-(-5)+/- sqrt((-5^2)-4(70)(-15))/(2*70)

the +/- means you do the equation once with a positive sign and a second time with a negative sign there. Does that make sense?

Answer 9

yes thanks

Answer 10

You're welcome :D

Answer 11

70z^2 - 5z - 15
5(14z^2-z - 3)
5(14z^2-7z + 6z -3)
5(7z(2z-1)+3(2z-1))
5((2z-1)(7z+3))

z = 1/2, -3/7

Answer 12

I hate when that happens.

Answer 13

In my opinion Quadratic is easier :P

Answer 14

No, you solved it before I did. I started solving it after you solved it already.

Answer 15

Yes, i did. Look wayy up there.

Answer 16

Yeah, but mine looks prettier

Answer 17

Okay. :)

Answer 18

And I solved it by factoring, showing that I'm not dependent on the quadratic formula, which to me, is always a last resort ( a resort I never use)

Answer 19

I'm the master of solving algebra without quad formulas, FOIL, or traditional substitution or elimination methods

Answer 20

sider 1993, any thoughts or comments?

Answer 21

well if I had not taken calculus in high school before this college course of algebra I would have gotten lost with the quadratic equation method because of right now we are using what hero used.