What are the solutions to the equation 3(4n-1)^2+1=-16 ?

The solutions to the equation 3(4n-1)^2+1=-16 are n= 1/4+i(sqrt(51))/(12),n= 1/4-i(sqrt(51))/(12)

Find the zeros of 3(4n-1)^2+1=-16

The zeros of 3(4n-1)^2+1=-16 are n= 1/4+i(sqrt(51))/(12),n= 1/4-i(sqrt(51))/(12)

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3 (4 n - 1)^{2} + 1 = - 16

n = \frac{1}{4} + i \frac{51}{12}, n = \frac{1}{4} - i \frac{51}{12}

Solution steps

3 (4 n - 1)^{2} + 1 = - 16

Move

1

to the right side

3 (4 n - 1)^{2} = - 17

Divide both sides by

3

(4 n - 1)^{2} = - \frac{17}{3}

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

4 n - 1 = - \frac{17}{3} : n = \frac{1}{4} + i \frac{51}{12}

Solve

4 n - 1 = - - \frac{17}{3} : n = \frac{1}{4} - i \frac{51}{12}

The solutions to the quadratic equation are:

n = \frac{1}{4} + i \frac{51}{12}, n = \frac{1}{4} - i \frac{51}{12}

7 n^{2} - 2 n - 36 = 6 n^{2} - 11 n

(x + 1)^{2} + (x - 3)^{2} = 15

8 = 2 x^{2}

2 x^{2} + 10 x + 3 = 0

5^{2} = 2^{2} + x^{2}

What are the solutions to the equation 3(4n-1)^2+1=-16 ?
The solutions to the equation 3(4n-1)^2+1=-16 are n= 1/4+i(sqrt(51))/(12),n= 1/4-i(sqrt(51))/(12)
Find the zeros of 3(4n-1)^2+1=-16
The zeros of 3(4n-1)^2+1=-16 are n= 1/4+i(sqrt(51))/(12),n= 1/4-i(sqrt(51))/(12)