What are the solutions to the equation 81n^2+33=42 ?

The solutions to the equation 81n^2+33=42 are n= 1/3 ,n=-1/3

Find the zeros of 81n^2+33=42

The zeros of 81n^2+33=42 are n= 1/3 ,n=-1/3

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81 n^{2} + 33 = 42

n = \frac{1}{3}, n = - \frac{1}{3}

Decimal Notation

n = 0.33333 \dots, n = - 0.33333 \dots

Solution steps

81 n^{2} + 33 = 42

Move

33

to the right side

81 n^{2} = 9

Divide both sides by

81

n^{2} = \frac{1}{9}

For

x^{2} = f (a)

the solutions are

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

n = \frac{1}{9}, n = - \frac{1}{9}

\frac{1}{9} = \frac{1}{3}

- \frac{1}{9} = - \frac{1}{3}

n = \frac{1}{3}, n = - \frac{1}{3}

27 - \frac{2}{3}

16 x^{7} y^{4}

x^{3} + 2 x^{2} - 8 x

\frac{( x ^{2} - 10 x + 24 )}{x ^{2} - 1} \leq 0

\frac{8 x ^{6} y ^{- 10} z ^{35}}{20 x ^{- 2} y ^{4} z ^{5}}

What are the solutions to the equation 81n^2+33=42 ?
The solutions to the equation 81n^2+33=42 are n= 1/3 ,n=-1/3
Find the zeros of 81n^2+33=42
The zeros of 81n^2+33=42 are n= 1/3 ,n=-1/3