Math, asked by ashu1232709, 11 months ago

pleave solve using quadratic equation​

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Answered by rishikapaul21
3

4 \sqrt{ 3}y ^{2}  + 5y - 2 \sqrt{3 = 0}

4 \sqrt{3} y ^{2}  + (8 - 3)y - 2 \sqrt{3}  = 0

4 \sqrt{3} y ^{2}  + 8y - 3y - 2 \sqrt{3}  = 0

4y( \sqrt{3} y + 2) -  \sqrt{3} ( \sqrt{3}y + 2)

( \sqrt{3} y + 2)(4y -  \sqrt{3}) = 0

either

 \sqrt{3} y + 2 = 0

y =  \frac{ - 2}{ \sqrt{3} }

or

4y -  \sqrt{3}  = 0

y =  \frac{ \sqrt{3} }{4}


rishikapaul21: yep
rishikapaul21: definition of chemical reactions
pkhasa16: Ok.
pkhasa16: I have written but it doesnot apper.
pkhasa16: Do you have any more questions tell all...
pkhasa16: I will write them all a put a question then..
pkhasa16: So ask the question if u want..
pkhasa16: Check in my below answer...
pkhasa16: Is it ok???
rishikapaul21: thnx bro
Answered by pkhasa16
6

Answer:

Step-by-step explanation:

JAI mahakal .

Check in the attachment....Ram ram.

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