What are the solutions to the equation (2x+8)^2=-29 ?

The solutions to the equation (2x+8)^2=-29 are x=-4+i(sqrt(29))/2 ,x=-4-i(sqrt(29))/2

Find the zeros of (2x+8)^2=-29

The zeros of (2x+8)^2=-29 are x=-4+i(sqrt(29))/2 ,x=-4-i(sqrt(29))/2

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(2 x + 8)^{2} = - 29

x = - 4 + i \frac{29}{2}, x = - 4 - i \frac{29}{2}

Solution steps

(2 x + 8)^{2} = - 29

For

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

the solutions are

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

Solve

2 x + 8 = - 29 : x = - 4 + i \frac{29}{2}

Solve

2 x + 8 = - - 29 : x = - 4 - i \frac{29}{2}

The solutions to the quadratic equation are:

x = - 4 + i \frac{29}{2}, x = - 4 - i \frac{29}{2}

4 x^{2} - 16 x + 12 = - 9 x + 9

227 - 0.25 q^{2} = 2 + 2 q^{2}

x^{2} + x - 20 = 10

- a^{2} - 3 a - 13 = - 2 a^{2}

y^{2} + 9 y + 3 = - 4 y^{2}

What are the solutions to the equation (2x+8)^2=-29 ?
The solutions to the equation (2x+8)^2=-29 are x=-4+i(sqrt(29))/2 ,x=-4-i(sqrt(29))/2
Find the zeros of (2x+8)^2=-29
The zeros of (2x+8)^2=-29 are x=-4+i(sqrt(29))/2 ,x=-4-i(sqrt(29))/2