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

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

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

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

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

x = - \frac{1}{2} + i \frac{1}{2}, x = - \frac{1}{2} - i \frac{1}{2}

Solution steps

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

For

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

the solutions are

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

Solve

2 x + 1 = - 1 : x = - \frac{1}{2} + i \frac{1}{2}

Solve

2 x + 1 = - - 1 : x = - \frac{1}{2} - i \frac{1}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{1}{2}, x = - \frac{1}{2} - i \frac{1}{2}

8 x + 4 + 3 x + 9 = 180

0 = - 0.6 t^{2} + 30 t

m^{2} - 4 n^{2}

5 x + 5 = 2 x + 2

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