What are the solutions to the equation (2x+5)^2=-4 ?

The solutions to the equation (2x+5)^2=-4 are x=-5/2+i,x=-5/2-i

Find the zeros of (2x+5)^2=-4

The zeros of (2x+5)^2=-4 are x=-5/2+i,x=-5/2-i

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(2 x + 5)^{2} = - 4

x = - \frac{5}{2} + i, x = - \frac{5}{2} - i

Solution steps

(2 x + 5)^{2} = - 4

For

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

the solutions are

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

Solve

2 x + 5 = - 4 : x = - \frac{5}{2} + i

Solve

2 x + 5 = - - 4 : x = - \frac{5}{2} - i

The solutions to the quadratic equation are:

x = - \frac{5}{2} + i, x = - \frac{5}{2} - i

(x - 4) (2 x - 3) = 12

4 x^{2} = 24 x - 35

x, f = x^{2} + 2 x y + y^{2}

y, y^{2} - 2 x y^{2} = x^{3} y + 5 x^{2} y^{2} - y

(x + 2)^{2} = 12 (x + 2)

What are the solutions to the equation (2x+5)^2=-4 ?
The solutions to the equation (2x+5)^2=-4 are x=-5/2+i,x=-5/2-i
Find the zeros of (2x+5)^2=-4
The zeros of (2x+5)^2=-4 are x=-5/2+i,x=-5/2-i