What are the solutions to the equation 3n-6=18n^2 ?

The solutions to the equation 3n-6=18n^2 are n= 1/12+i(sqrt(47))/(12),n= 1/12-i(sqrt(47))/(12)

Find the zeros of 3n-6=18n^2

The zeros of 3n-6=18n^2 are n= 1/12+i(sqrt(47))/(12),n= 1/12-i(sqrt(47))/(12)

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3 n - 6 = 18 n^{2}

n = \frac{1}{12} + i \frac{47}{12}, n = \frac{1}{12} - i \frac{47}{12}

Solution steps

3 n - 6 = 18 n^{2}

Switch sides

18 n^{2} = 3 n - 6

Move

6

to the left side

18 n^{2} + 6 = 3 n

Move

3 n

to the left side

18 n^{2} + 6 - 3 n = 0

Write in the standard form

a x^{2} + b x + c = 0

18 n^{2} - 3 n + 6 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 18 \cdot 6}{2 \cdot 18}

Simplify

(- 3)^{2} - 4 \cdot 18 \cdot 6 : 347 i

n_{1, 2} = \frac{- ( - 3 ) \pm 3 47 i}{2 \cdot 18}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 3 47 i}{2 \cdot 18}, n_{2} = \frac{- ( - 3 ) - 3 47 i}{2 \cdot 18}

n = \frac{- ( - 3 ) + 3 47 i}{2 \cdot 18} : \frac{1}{12} + i \frac{47}{12}

n = \frac{- ( - 3 ) - 3 47 i}{2 \cdot 18} : \frac{1}{12} - i \frac{47}{12}

The solutions to the quadratic equation are:

n = \frac{1}{12} + i \frac{47}{12}, n = \frac{1}{12} - i \frac{47}{12}

3^{2} = x^{2} + 1^{2}

x^{2} + 3 x (x - 12) = 0

3 x^{2} - 12 x + 5

2 x^{2} - 3 x - 20 = x^{2} + 34

2 x^{2} + 2 = 10

What are the solutions to the equation 3n-6=18n^2 ?
The solutions to the equation 3n-6=18n^2 are n= 1/12+i(sqrt(47))/(12),n= 1/12-i(sqrt(47))/(12)
Find the zeros of 3n-6=18n^2
The zeros of 3n-6=18n^2 are n= 1/12+i(sqrt(47))/(12),n= 1/12-i(sqrt(47))/(12)