What are the solutions to the equation (4x+1)^2=3 ?

The solutions to the equation (4x+1)^2=3 are x=(sqrt(3)-1)/4 ,x=(-sqrt(3)-1)/4

Find the zeros of (4x+1)^2=3

The zeros of (4x+1)^2=3 are x=(sqrt(3)-1)/4 ,x=(-sqrt(3)-1)/4

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(4 x + 1)^{2} = 3

x = \frac{3 - 1}{4}, x = \frac{- 3 - 1}{4}

Decimal Notation

x = 0.18301 \dots, x = - 0.68301 \dots

Solution steps

(4 x + 1)^{2} = 3

For

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

the solutions are

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

Solve

4 x + 1 = 3 : x = \frac{3 - 1}{4}

Solve

4 x + 1 = - 3 : x = \frac{- 3 - 1}{4}

The solutions to the quadratic equation are:

x = \frac{3 - 1}{4}, x = \frac{- 3 - 1}{4}

2 (\frac{3}{2})

- \frac{1}{2} \cdot \frac{1}{3}

(- 64)^{- \frac{1}{3}}

\frac{3}{5} b - \frac{1}{2} b \geq \frac{1}{5} b - \frac{1}{2}

4 x^{4} y^{6} - 25

What are the solutions to the equation (4x+1)^2=3 ?
The solutions to the equation (4x+1)^2=3 are x=(sqrt(3)-1)/4 ,x=(-sqrt(3)-1)/4
Find the zeros of (4x+1)^2=3
The zeros of (4x+1)^2=3 are x=(sqrt(3)-1)/4 ,x=(-sqrt(3)-1)/4