What are the solutions to the equation -3(y+4)^2=-48 ?

The solutions to the equation -3(y+4)^2=-48 are y=0,y=-8

Find the zeros of -3(y+4)^2=-48

The zeros of -3(y+4)^2=-48 are y=0,y=-8

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- 3 (y + 4)^{2} = - 48

y = 0, y = - 8

Solution steps

- 3 (y + 4)^{2} = - 48

Divide both sides by

- 3

(y + 4)^{2} = 16

For

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

the solutions are

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

Solve

y + 4 = 16 : y = 0

Solve

y + 4 = - 16 : y = - 8

The solutions to the quadratic equation are:

y = 0, y = - 8

(x - 3) (x^{2} + 10 x + 25) \geq 0

(3 x + 2)^{2} + 8 (3 x + 2) + 12

2 x - 4 = 2 x

\frac{2}{3} - \frac{1}{2} + \frac{2}{4}

a^{2} - 9 a + 20

What are the solutions to the equation -3(y+4)^2=-48 ?
The solutions to the equation -3(y+4)^2=-48 are y=0,y=-8
Find the zeros of -3(y+4)^2=-48
The zeros of -3(y+4)^2=-48 are y=0,y=-8