What are the solutions to the equation 3m^2+13=m^2+397 ?

The solutions to the equation 3m^2+13=m^2+397 are m=8sqrt(3),m=-8sqrt(3)

Find the zeros of 3m^2+13=m^2+397

The zeros of 3m^2+13=m^2+397 are m=8sqrt(3),m=-8sqrt(3)

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3 m^{2} + 13 = m^{2} + 397

m = 83, m = - 83

Solution steps

3 m^{2} + 13 = m^{2} + 397

Move

13

to the right side

3 m^{2} = m^{2} + 384

Move

m^{2}

to the left side

2 m^{2} = 384

Divide both sides by

2

m^{2} = 192

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

m = 192, m = - 192

192 = 83

- 192 = - 83

m = 83, m = - 83

2 x^{2} y - 5 x^{2} - 2 y + 5

\frac{3}{4} = \frac{n}{12} n

\frac{1}{2} f - 4 = 1

- \frac{23}{2} (\frac{1}{100 0 ^{2}}) + \frac{23}{2}

6

What are the solutions to the equation 3m^2+13=m^2+397 ?
The solutions to the equation 3m^2+13=m^2+397 are m=8sqrt(3),m=-8sqrt(3)
Find the zeros of 3m^2+13=m^2+397
The zeros of 3m^2+13=m^2+397 are m=8sqrt(3),m=-8sqrt(3)