What are the solutions to the equation 3(y+1)^2-5=43 ?

The solutions to the equation 3(y+1)^2-5=43 are y=3,y=-5

Find the zeros of 3(y+1)^2-5=43

The zeros of 3(y+1)^2-5=43 are y=3,y=-5

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3 (y + 1)^{2} - 5 = 43

y = 3, y = - 5

Solution steps

3 (y + 1)^{2} - 5 = 43

Move

5

to the right side

3 (y + 1)^{2} = 48

Divide both sides by

3

(y + 1)^{2} = 16

For

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

the solutions are

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

Solve

y + 1 = 16 : y = 3

Solve

y + 1 = - 16 : y = - 5

The solutions to the quadratic equation are:

y = 3, y = - 5

4 \cdot 1.5

\frac{( y - 2 ) ^{2}}{3} = - y + 8

- 6 \leq x \leq 4

8 x + 2 > 10

\frac{∣ j ∣}{5} + 1 = - 4

What are the solutions to the equation 3(y+1)^2-5=43 ?
The solutions to the equation 3(y+1)^2-5=43 are y=3,y=-5
Find the zeros of 3(y+1)^2-5=43
The zeros of 3(y+1)^2-5=43 are y=3,y=-5