What are the solutions to the equation (y+2)^2+152=0 ?

The solutions to the equation (y+2)^2+152=0 are y=2sqrt(38)i-2,y=-2sqrt(38)i-2

Find the zeros of (y+2)^2+152=0

The zeros of (y+2)^2+152=0 are y=2sqrt(38)i-2,y=-2sqrt(38)i-2

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(y + 2)^{2} + 152 = 0

y = 238 i - 2, y = - 238 i - 2

Solution steps

(y + 2)^{2} + 152 = 0

Move

152

to the right side

(y + 2)^{2} = - 152

For

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

the solutions are

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

Solve

y + 2 = - 152 : y = 238 i - 2

Solve

y + 2 = - - 152 : y = - 238 i - 2

The solutions to the quadratic equation are:

y = 238 i - 2, y = - 238 i - 2

- x^{2} + 100 x - 2500 = 0

c^{2} - 9 = 0

\frac{x ^{2}}{2} + \frac{x}{2} = 10

- 3 x = 5 x^{2}

9 x^{2} = - 6 x - 1

What are the solutions to the equation (y+2)^2+152=0 ?
The solutions to the equation (y+2)^2+152=0 are y=2sqrt(38)i-2,y=-2sqrt(38)i-2
Find the zeros of (y+2)^2+152=0
The zeros of (y+2)^2+152=0 are y=2sqrt(38)i-2,y=-2sqrt(38)i-2