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

The solutions to the equation (2x+3)^2=-4 are x=-3/2+i,x=-3/2-i

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

The zeros of (2x+3)^2=-4 are x=-3/2+i,x=-3/2-i

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

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

Solution steps

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

For

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

the solutions are

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

Solve

2 x + 3 = - 4 : x = - \frac{3}{2} + i

Solve

2 x + 3 = - - 4 : x = - \frac{3}{2} - i

The solutions to the quadratic equation are:

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

4 t^{2} + 2 t - (3 + 3) = 0

x^{2} + 2 = - 4 x

c^{2} - 7 c + 3 = - 2 c

2^{2} + x^{2} = 1 0^{2}

x^{2} + 2 x + 30 = 0

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