What are the solutions to the equation 25(y+5)^2=81 ?

The solutions to the equation 25(y+5)^2=81 are y=-16/5 ,y=-34/5

Find the zeros of 25(y+5)^2=81

The zeros of 25(y+5)^2=81 are y=-16/5 ,y=-34/5

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25 (y + 5)^{2} = 81

y = - \frac{16}{5}, y = - \frac{34}{5}

Decimal Notation

y = - 3.2, y = - 6.8

Solution steps

25 (y + 5)^{2} = 81

Divide both sides by

25

(y + 5)^{2} = \frac{81}{25}

For

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

the solutions are

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

Solve

y + 5 = \frac{81}{25} : y = - \frac{16}{5}

Solve

y + 5 = - \frac{81}{25} : y = - \frac{34}{5}

The solutions to the quadratic equation are:

y = - \frac{16}{5}, y = - \frac{34}{5}

7 x^{2} + 27 x - 4 = 0

x^{2} + 5 x + 4 = 9

x^{2} + \frac{4}{9} x

x (2 x + 1) = 3

1156 = x^{2}

What are the solutions to the equation 25(y+5)^2=81 ?
The solutions to the equation 25(y+5)^2=81 are y=-16/5 ,y=-34/5
Find the zeros of 25(y+5)^2=81
The zeros of 25(y+5)^2=81 are y=-16/5 ,y=-34/5