Need help on Quadratic equations please!
The equation is on the comments.

- andriod09

Need help on Quadratic equations please!
The equation is on the comments.

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- andriod09

\[4x^{2}+9x=0\]
i know that the answer is:
\[x=\frac{ -9\pm \sqrt{81-4(4)(2)} }{ 8 }\]

- anonymous

Much easier to solve your problem would be to factor out an x,\[x(4x+9)=0\]now you have two solutions,\[x=0\]and\[4x+9=0\]\[4x=-9\]\[x=-\frac{9}{4}\]and you get

- anonymous

The equation is correct. But you put innocent "2" instead of "0" for c.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- andriod09

huh? the thing messed up, give me a sec

- anonymous

Well, then you can do the following:\[x_{1/2}=\frac{-9 \pm \sqrt{9^2-4*4*0}}{2*4}\]simplifying that you get,\[x_{1/2}=\frac{ -9 \pm 9 }{ 8 }\]Compute the rest

- andriod09

I have to use:
\[ax^{2}+bx+c=0\]
and i have to use:
\[x=\frac{ -b\pm \sqrt{b^{2}-4ac} }{ 2a }\]

- anonymous

Yes.. that is what I have used. Just make c=0, and you will get the same result as I have.

- andriod09

but c = 2... thats what the answer says.

- anonymous

Well then you have given us a wrong problem\[4x^2+9x=0\]does not have a c in it..

- andriod09

my bad... /facepalm
its:
\[4x^{2}+9x+2=0\]

- andriod09

- anonymous

Hehe, no worries. Now you have to use the formula you input\[x_{1/2}=\frac{-9 \pm \sqrt{9^2-4*4*2}}{2*4}\]compute the values inside the square root first\[x_{1/2}=\frac{-9 \pm \sqrt{81- 32}}{8}\]and complete calculation\[x_{1/2}=\frac{-9 \pm 7}{8}\] and you get,\[x_1= -2\]\[x_2=-\frac{1}{4}\]

- andriod09

where did you get \[x _{1/2} \]

- anonymous

That just means that there are two solutions we will get from the equation. Its just mathematical notation. I am used to writing it like that. Don't worry, you can write just x, it will be fine.

- andriod09

im only in high school. :/ :L

- anonymous

Yeah I figured. The quadratic equations are studied extensively in high school. You're doing good tho. Might want to reconsider your profile picture heh ;)

- andriod09

it says i don't hate math. open my profile and look at the white! everyone says that. lol do you not see the white scribble on the \[^{don't}\]
I hate math

- anonymous

Oh.. then I owe you an apology. XD.

- andriod09

Its okay. I need to make it bigger anyway. lol. Thanks tho.

- anonymous

You're welcome man. Anytime!

- andriod09

I gave you a medal. dont for get to give one back~ ;)

- anonymous

Hey, my bad, my first day over here.

- andriod09

what do you mean?

- anonymous

This is my first day in open study, so I'm still getting used to the medals and stuff.

- andriod09

how would i set up:
\[3x^{2}+11x+5=0\] and oh. i didn't know

- anonymous

Same principle, replace a=3, b=11, and c=5.

- andriod09

i mean in the fraction form

Looking for something else?

Not the answer you are looking for? Search for more explanations.