What are the solutions to the equation (5y+4)^2+100=0 ?

The solutions to the equation (5y+4)^2+100=0 are y=-4/5+2i,y=-4/5-2i

Find the zeros of (5y+4)^2+100=0

The zeros of (5y+4)^2+100=0 are y=-4/5+2i,y=-4/5-2i

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(5 y + 4)^{2} + 100 = 0

y = - \frac{4}{5} + 2 i, y = - \frac{4}{5} - 2 i

Solution steps

(5 y + 4)^{2} + 100 = 0

Move

100

to the right side

(5 y + 4)^{2} = - 100

For

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

the solutions are

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

Solve

5 y + 4 = - 100 : y = - \frac{4}{5} + 2 i

Solve

5 y + 4 = - - 100 : y = - \frac{4}{5} - 2 i

The solutions to the quadratic equation are:

y = - \frac{4}{5} + 2 i, y = - \frac{4}{5} - 2 i

7 = \frac{1}{3} (8 x + 6)

3 x + 7 = 19

2 x + 1 < 4 x + 5

2 i (7 - i)

x + 2 = - 12

What are the solutions to the equation (5y+4)^2+100=0 ?
The solutions to the equation (5y+4)^2+100=0 are y=-4/5+2i,y=-4/5-2i
Find the zeros of (5y+4)^2+100=0
The zeros of (5y+4)^2+100=0 are y=-4/5+2i,y=-4/5-2i