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Answer :

Answer:

=

βˆ’

7

3

=

βˆ’

7

Step-by-step explanation:

=

βˆ’

Β±

2

βˆ’

4

√

2

x={b}}^{2}-4{a}}{2{a}}}

x=2aβˆ’bΒ±b2βˆ’4ac​​

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

3

2

+

2

8

+

4

9

=

0

3x^{2}+28x+49=0

3x2+28x+49=0

=

3

a={3}}

a=3

=

2

8

b={28}}

b=28

=

4

9

c={129}{49}}

c=49

=

βˆ’

2

8

Β±

2

8

2

βˆ’

4

β‹…

3

β‹…

4

9

√

2

β‹…

3

=

βˆ’

2

8

Β±

1

4 over

6

=

βˆ’

2

8

+

1

4

6

x={-28+14}{6}

x=6βˆ’28+14​

=

βˆ’

2

8

βˆ’

1

4

=

βˆ’

7

3

x=-{7}{3}

x=βˆ’37​

=

βˆ’

7

=

βˆ’

7

3

=

βˆ’

7

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