What are the solutions to the equation (7x+1)^2=5 ?

The solutions to the equation (7x+1)^2=5 are x=(sqrt(5)-1)/7 ,x=(-sqrt(5)-1)/7

Find the zeros of (7x+1)^2=5

The zeros of (7x+1)^2=5 are x=(sqrt(5)-1)/7 ,x=(-sqrt(5)-1)/7

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(7 x + 1)^{2} = 5

x = \frac{5 - 1}{7}, x = \frac{- 5 - 1}{7}

Decimal Notation

x = 0.17658 \dots, x = - 0.46229 \dots

Solution steps

(7 x + 1)^{2} = 5

For

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

the solutions are

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

Solve

7 x + 1 = 5 : x = \frac{5 - 1}{7}

Solve

7 x + 1 = - 5 : x = \frac{- 5 - 1}{7}

The solutions to the quadratic equation are:

x = \frac{5 - 1}{7}, x = \frac{- 5 - 1}{7}

\frac{1}{2} - \frac{2}{5}

\frac{x}{2} + \frac{5 x}{4} = \frac{x}{12}

- 1 + \frac{2}{3}

\frac{7 x}{5} - \frac{3 x}{10} = 2 (x + \frac{9}{2})

1 0^{2} + a^{2} = 1 5^{2}

What are the solutions to the equation (7x+1)^2=5 ?
The solutions to the equation (7x+1)^2=5 are x=(sqrt(5)-1)/7 ,x=(-sqrt(5)-1)/7
Find the zeros of (7x+1)^2=5
The zeros of (7x+1)^2=5 are x=(sqrt(5)-1)/7 ,x=(-sqrt(5)-1)/7