What are the solutions to the equation (2x+3)^2+25=0 ?

The solutions to the equation (2x+3)^2+25=0 are x=-3/2+i 5/2 ,x=-3/2-i 5/2

Find the zeros of (2x+3)^2+25=0

The zeros of (2x+3)^2+25=0 are x=-3/2+i 5/2 ,x=-3/2-i 5/2

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(2 x + 3)^{2} + 25 = 0

x = - \frac{3}{2} + i \frac{5}{2}, x = - \frac{3}{2} - i \frac{5}{2}

Solution steps

(2 x + 3)^{2} + 25 = 0

Move

25

to the right side

(2 x + 3)^{2} = - 25

For

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

the solutions are

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

Solve

2 x + 3 = - 25 : x = - \frac{3}{2} + i \frac{5}{2}

Solve

2 x + 3 = - - 25 : x = - \frac{3}{2} - i \frac{5}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{5}{2}, x = - \frac{3}{2} - i \frac{5}{2}

(\frac{1}{4})^{1}

- 5 x + 12 = - 12 x - 12

\frac{( \frac{a}{b} )}{c} + \frac{d}{( \frac{e}{f} )}

27 x^{3} - 12 x

\frac{a}{2} \leq 2

What are the solutions to the equation (2x+3)^2+25=0 ?
The solutions to the equation (2x+3)^2+25=0 are x=-3/2+i 5/2 ,x=-3/2-i 5/2
Find the zeros of (2x+3)^2+25=0
The zeros of (2x+3)^2+25=0 are x=-3/2+i 5/2 ,x=-3/2-i 5/2