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

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

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

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

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(2 y - 3)^{2} + 25 = 0

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

Solution steps

(2 y - 3)^{2} + 25 = 0

Move

25

to the right side

(2 y - 3)^{2} = - 25

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

0 = 12 (x - 3)^{2}

12 x^{2} - 32 x + 16 = 0

7^{2} + b^{2} = 8^{2}

- 4 x^{2} + x - 9 = 0

m^{2} = 144

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