Question

write each equation in standard form then solve
21x^2 +5x-35=3x^2-4x

Answer 1

? where is the question

Answer 2

you can take this one @texaschic101 XD

Answer 3

Have you considered moving everything to the left-hand side, combining like terms, and then solving?

Answer 4

so -3x^2-4x?

Answer 5

or +4x i mean

Answer 6

Not close. Where did the "=" go?

Answer 7

21x^2 + 5x - 35 = 3x^2 - 4x
21x^2 - 3x^2 + 5x + 4x - 35 = 0
17x^2 + 9x - 35 = 0

now use the quadratic formula
http://www.purplemath.com/modules/quadform.htm

Answer 8

ok but how did you go from 21x^2-3x^2+5x+4x-35=0 to 17x^2=9x-35=0?

Answer 9

21-3 = 18

Answer 10

ok then i multiply 18 times -35?

Answer 11

Why would you do that? Factor or use the quadratic formula.

Answer 12

or do i just do the x method

Answer 13

There are many names for many things. If that is what you mean for finding the factors, then yes. If that is what you mean for identifying your mutant powers, then no.

Answer 14

I can't find the factors that add to 9 but multiply to -35

Answer 15

Super. Have you considered the quadratic formula? Maybe no such factors exist.

Answer 16

i'm so confused

Answer 17

21x^2 +5x-35=3x^2-4x

18x^2 + 9x - 35 = 0

9^2 - 4(18)(-35) = 81 + 2520 = 2601

\(\sqrt{2601} = 51\) - And this is where we feel bad because it was a perfect square and we should have been able to factor it. Since we didn't, we have...

\(x = \dfrac{9 \pm 51}{42}\)

That should fill you in on how to factor it.

Answer 18

Just confused me even more but nevermind i'll just try to ask my teacher for more help