What are the solutions to the equation (4x+1)^2=7 ?

The solutions to the equation (4x+1)^2=7 are x=(sqrt(7)-1)/4 ,x=(-sqrt(7)-1)/4

Find the zeros of (4x+1)^2=7

The zeros of (4x+1)^2=7 are x=(sqrt(7)-1)/4 ,x=(-sqrt(7)-1)/4

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

x = \frac{7 - 1}{4}, x = \frac{- 7 - 1}{4}

Decimal Notation

x = 0.41143 \dots, x = - 0.91143 \dots

Solution steps

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

For

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

the solutions are

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

Solve

4 x + 1 = 7 : x = \frac{7 - 1}{4}

Solve

4 x + 1 = - 7 : x = \frac{- 7 - 1}{4}

The solutions to the quadratic equation are:

x = \frac{7 - 1}{4}, x = \frac{- 7 - 1}{4}

3 (x^{2} + 7 x + 10) = 0

- 3 x^{2} + 7 x = - 5

(x^{2} + 3 x - 1)^{2} < 9

r^{2} - 36 = 0

(x + h)^{2} - x^{2}

What are the solutions to the equation (4x+1)^2=7 ?
The solutions to the equation (4x+1)^2=7 are x=(sqrt(7)-1)/4 ,x=(-sqrt(7)-1)/4
Find the zeros of (4x+1)^2=7
The zeros of (4x+1)^2=7 are x=(sqrt(7)-1)/4 ,x=(-sqrt(7)-1)/4