What are the solutions to the equation (2x+5)^2+1=0 ?

The solutions to the equation (2x+5)^2+1=0 are x=-5/2+i 1/2 ,x=-5/2-i 1/2

Find the zeros of (2x+5)^2+1=0

The zeros of (2x+5)^2+1=0 are x=-5/2+i 1/2 ,x=-5/2-i 1/2

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(2 x + 5)^{2} + 1 = 0

x = - \frac{5}{2} + i \frac{1}{2}, x = - \frac{5}{2} - i \frac{1}{2}

Solution steps

(2 x + 5)^{2} + 1 = 0

Move

1

to the right side

(2 x + 5)^{2} = - 1

For

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

the solutions are

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

Solve

2 x + 5 = - 1 : x = - \frac{5}{2} + i \frac{1}{2}

Solve

2 x + 5 = - - 1 : x = - \frac{5}{2} - i \frac{1}{2}

The solutions to the quadratic equation are:

x = - \frac{5}{2} + i \frac{1}{2}, x = - \frac{5}{2} - i \frac{1}{2}

x, f (x + 1) = 4 x^{2} + 2 x - 4

x^{2} - 16 x = - 80

2 x^{2} - 196 = x^{2}

x^{2} - 16 x = - 62

x, x y (x + y) = 25

What are the solutions to the equation (2x+5)^2+1=0 ?
The solutions to the equation (2x+5)^2+1=0 are x=-5/2+i 1/2 ,x=-5/2-i 1/2
Find the zeros of (2x+5)^2+1=0
The zeros of (2x+5)^2+1=0 are x=-5/2+i 1/2 ,x=-5/2-i 1/2