What are the solutions to the equation -3(2x+1)^3=-192 ?

The solutions to the equation -3(2x+1)^3=-192 are x= 3/2 ,x=-3/2+sqrt(3)i,x=-3/2-sqrt(3)i

Find the zeros of -3(2x+1)^3=-192

The zeros of -3(2x+1)^3=-192 are x= 3/2 ,x=-3/2+sqrt(3)i,x=-3/2-sqrt(3)i

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- 3 (2 x + 1)^{3} = - 192

x = \frac{3}{2}, x = - \frac{3}{2} + 3 i, x = - \frac{3}{2} - 3 i

Solution steps

- 3 (2 x + 1)^{3} = - 192

Divide both sides by

- 3

(2 x + 1)^{3} = 64

For

g^{3} (x) = f (a)

the solutions are

Solve

The solutions are

x = \frac{3}{2}, x = - \frac{3}{2} + 3 i, x = - \frac{3}{2} - 3 i

(x^{2} - x + 1)^{2} = 0

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

x^{4} + 8 x^{3} + 17 x^{2} + 8 x + 16 = 0

x^{4} + 12 x + 3 = 0

0 = 4 x^{3} - 3 x^{2} + 1

What are the solutions to the equation -3(2x+1)^3=-192 ?
The solutions to the equation -3(2x+1)^3=-192 are x= 3/2 ,x=-3/2+sqrt(3)i,x=-3/2-sqrt(3)i
Find the zeros of -3(2x+1)^3=-192
The zeros of -3(2x+1)^3=-192 are x= 3/2 ,x=-3/2+sqrt(3)i,x=-3/2-sqrt(3)i