What are the solutions to the equation (2x+7)^2+4=0 ?

The solutions to the equation (2x+7)^2+4=0 are x=-7/2+i,x=-7/2-i

Find the zeros of (2x+7)^2+4=0

The zeros of (2x+7)^2+4=0 are x=-7/2+i,x=-7/2-i

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(2 x + 7)^{2} + 4 = 0

x = - \frac{7}{2} + i, x = - \frac{7}{2} - i

Solution steps

(2 x + 7)^{2} + 4 = 0

Move

4

to the right side

(2 x + 7)^{2} = - 4

For

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

the solutions are

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

Solve

2 x + 7 = - 4 : x = - \frac{7}{2} + i

Solve

2 x + 7 = - - 4 : x = - \frac{7}{2} - i

The solutions to the quadratic equation are:

x = - \frac{7}{2} + i, x = - \frac{7}{2} - i

3 x^{2} + 9 x = 0

2 k^{2} + 13 k + 18

x^{2} + 4 = 8

12 a^{5} b^{2} - 36 a^{4} b^{3} - 6 a^{2} b^{2}

y^{2} - y

What are the solutions to the equation (2x+7)^2+4=0 ?
The solutions to the equation (2x+7)^2+4=0 are x=-7/2+i,x=-7/2-i
Find the zeros of (2x+7)^2+4=0
The zeros of (2x+7)^2+4=0 are x=-7/2+i,x=-7/2-i