What are the solutions to the equation (5y+2)^2-112=0 ?

The solutions to the equation (5y+2)^2-112=0 are y=(4sqrt(7)-2)/5 ,y=(-4sqrt(7)-2)/5

Find the zeros of (5y+2)^2-112=0

The zeros of (5y+2)^2-112=0 are y=(4sqrt(7)-2)/5 ,y=(-4sqrt(7)-2)/5

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(5 y + 2)^{2} - 112 = 0

y = \frac{4 7 - 2}{5}, y = \frac{- 4 7 - 2}{5}

Decimal Notation

y = 1.71660 \dots, y = - 2.51660 \dots

Solution steps

(5 y + 2)^{2} - 112 = 0

Move

112

to the right side

(5 y + 2)^{2} = 112

For

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

the solutions are

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

Solve

5 y + 2 = 112 : y = \frac{4 7 - 2}{5}

Solve

5 y + 2 = - 112 : y = \frac{- 4 7 - 2}{5}

The solutions to the quadratic equation are:

y = \frac{4 7 - 2}{5}, y = \frac{- 4 7 - 2}{5}

(7^{- 2})^{6}

r^{2} - 4 r + 4 = 0

x^{5} - 2 x^{4} - x^{3} - 2 x^{2} - 20 x + 24 < 0

\frac{4 + i}{5 + i}

9 x^{2} - 9 x + 2 = 0

What are the solutions to the equation (5y+2)^2-112=0 ?
The solutions to the equation (5y+2)^2-112=0 are y=(4sqrt(7)-2)/5 ,y=(-4sqrt(7)-2)/5
Find the zeros of (5y+2)^2-112=0
The zeros of (5y+2)^2-112=0 are y=(4sqrt(7)-2)/5 ,y=(-4sqrt(7)-2)/5