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

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

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

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

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

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

Solution steps

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

Move

1

to the right side

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

Divide both sides by

3

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

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

- 0.5 x^{2} + 3.2 x - 2 = 0

3 x^{2} - 16 x

3 x^{2} - 15 x

r^{2} = 2 z - z^{2}

x, f = (2 m - 1) x^{2} - 10 x - m

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