Can someone help me:(?
Decide which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring.
-b b2 - 4ac 2a
Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation:
4x2 + 6x + 2 = 0 (2 points)

- anonymous

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- anonymous

this was another one i dint quite get

- pooja195

The "discriminant" gives this information. It is the square root part of the quadratic formula. \[\sqrt{b^2-4ac}\]

- pooja195

From here solve the equation

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## More answers

- pooja195

Start by listing your abc values
a=?
b=?
c=?

- pooja195

\[\huge~\rm~ax^2+bx+c=0\]

- anonymous

4x^2+6x++2=0

- anonymous

is that how it should be set up in order

- pooja195

yes but we need to figure out the values using
\[\huge~\rm~ax^2+bx+c=0\]
That equation can you list just the NUMBERS not the signs of variables

- anonymous

im confused

- arindameducationusc

@pooja195 after this, can you solve my question?
http://openstudy.com/study#/updates/55b787e4e4b0f0c1f13fd1a9

- pooja195

Ok anything ^2 is the a value anything with an "x" is the b value anything with no varible is the c value now list the numbers :)

- anonymous

so b^2 is A, -b is B, and C is 2a?

- pooja195

wut? o-o
wow you really are confused ok lets start from scratch
\[\huge~\rm~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
This is the quadratic formula
\[\huge~\rm~ax^2+bx+c=0\]
This is standered form
The discernment is The "discriminant" gives this information. It is the square root part of the quadratic formula. \[\sqrt{b^2-4ac}\]
We need to solve
\[\huge~\rm~4x^2 + 6x + 2 = 0\]
Notice how it all matches up with the standered form equation?
\[\huge~\rm~ax^2+bx+c=0\]

- anonymous

i know, i said that already

- pooja195

You have the numbers wrong
a=4
b=6
c=2

- anonymous

oh

- anonymous

then why do we need just the numbers

- pooja195

We need just the numbers so that we can solve using the discernment

- anonymous

ok

- arindameducationusc

@pooja195 I request you... okay after this

- pooja195

\[\sqrt{b^2-4ac}\]
So now we have our abc values
\[\huge~\rm~ \sqrt{6^2-4(4)(2)}\]
Solve that

- anonymous

6^2-8 under a square root idk what

- anonymous

i started with parenthesis and then multiplied 4 by (8)

- pooja195

ok lets do this step by step whats 6^2?

- anonymous

36

- pooja195

ok leave that aside whats -4 times 4 ?

- anonymous

-16

- pooja195

Ok good -16 x 2=?

- anonymous

-32

- pooja195

good whats 36-32?

- anonymous

4

- anonymous

The 4ac

- anonymous

4ac decides whether you get complex number or not.

- pooja195

good now whats the
\[\huge~\rm \sqrt4\]

- anonymous

2

- pooja195

@saseal you are confusing the asker we are already done with the problem.

- anonymous

oh ok

- pooja195

@kinserkara so whats your final answer?

- anonymous

sry about that

- anonymous

2a

- pooja195

its ok :)
@kinserkara now 2a it would just be x=2

- anonymous

so the final answer is x=2

- pooja195

yep

- anonymous

ok thanks

- pooja195

No problem :)

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