@whpalmer4 4x^2+4x=15. solve by quadratic formula. does this mean I use the x=-b= +/-sq.rt b2-4ac/2a or do i use the ax^2 +bx +c =0 method?

Question

@whpalmer4 4x^2+4x=15.  solve by quadratic formula.  does this mean I use the x=-b= +/-sq.rt b2-4ac/2a or do i use the ax^2 +bx +c =0 method?

anonymous · Answer

The first one$$x=\frac{ -b \pm \sqrt{b^{2}-4ac} }{ 2a }$$
is the quadratic formula

anonymous · Answer

thank you.  give me a minute to pen and paper this to see if i can get a correct answer

anonymous · Answer

No problem

anonymous · Answer

okay the closest to an answer I can get is (-4 +/- 16)/8.  not quite sure how to proceed to further simplify

anonymous · Answer

-20/8 or 12/8 then reduces to...-5/2 or 3/2, yes?  so x= -5/2 or 3/2?

anonymous · Answer

Quite right!

anonymous · Answer

thankee-sai.  your help is most appreciated :)

anonymous · Answer

No problem, you did all of the work :)

anonymous · Answer

i usually just need a little nudge to point me in the right direction.  i won't learn a thing if someone else does the work

anonymous · Answer

I agree and its great that you realize that, especially with math. Feel free to tag me anytime you need a little nudge lol

anonymous · Answer

thankee-sai and will do.  Blessed be!

anonymous · Answer

And you!

whpalmer4 · Answer

You have to put it in the form $ax^2+bx+c=0$ first in order to get the correct values of $a,b,c$ for the quadratic formula
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

whpalmer4 · Answer

$$x=-\frac{5}{2},\,x =\frac{3}{2}$$are the correct solutions once you do that.