What are the solutions to the equation (5y+23)^2=32 ?

The solutions to the equation (5y+23)^2=32 are y=(4sqrt(2)-23)/5 ,y=(-4sqrt(2)-23)/5

Find the zeros of (5y+23)^2=32

The zeros of (5y+23)^2=32 are y=(4sqrt(2)-23)/5 ,y=(-4sqrt(2)-23)/5

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(5 y + 23)^{2} = 32

y = \frac{4 2 - 23}{5}, y = \frac{- 4 2 - 23}{5}

Decimal Notation

y = - 3.46862 \dots, y = - 5.73137 \dots

Solution steps

(5 y + 23)^{2} = 32

For

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

the solutions are

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

Solve

5 y + 23 = 32 : y = \frac{4 2 - 23}{5}

Solve

5 y + 23 = - 32 : y = \frac{- 4 2 - 23}{5}

The solutions to the quadratic equation are:

y = \frac{4 2 - 23}{5}, y = \frac{- 4 2 - 23}{5}

lo g_{6} (216)

2 p + 3 - 1 \geq - (p - 3) - 2 (2 p + 4)

x^{2} = x

5 x^{2} + 2 x + 10 = 0

- \frac{1}{2} (2)^{- 1}

What are the solutions to the equation (5y+23)^2=32 ?
The solutions to the equation (5y+23)^2=32 are y=(4sqrt(2)-23)/5 ,y=(-4sqrt(2)-23)/5
Find the zeros of (5y+23)^2=32
The zeros of (5y+23)^2=32 are y=(4sqrt(2)-23)/5 ,y=(-4sqrt(2)-23)/5