Question

4n^2+1204n-48528 = 0 ...solve this pls

Answer 1

Step-by-step explanation:

=

−

±

2

−

4

√

2

n=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

n=2a−b±b2−4ac

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

4

2

+

1

2

0

4

−

4

8

5

2

8

=

0

4n^{2}+1204n-48528=0

4n2+1204n−48528=0

=

4

a={\color{#c92786}{4}}

a=4

=

1

2

0

4

b={\color{#e8710a}{1204}}

b=1204

=

−

4

8

5

2

8

c={\color{#129eaf}{-48528}}

c=−48528

=

−

1

2

0

4

±

1

2

0

4

2

−

4

⋅

4

(

−

4

8

5

2

8

)

√

2

⋅

4

Answer 2

Answer:

hiiii

I'm busy with my assignment now

Step-by-step explanation:

I'll ch.at with u later

bye

take care