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

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

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

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

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

x = \frac{3 - 1}{2}, x = \frac{- 3 - 1}{2}

Decimal Notation

x = 0.36602 \dots, x = - 1.36602 \dots

Solution steps

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

For

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

the solutions are

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

Solve

2 x + 1 = 3 : x = \frac{3 - 1}{2}

Solve

2 x + 1 = - 3 : x = \frac{- 3 - 1}{2}

The solutions to the quadratic equation are:

x = \frac{3 - 1}{2}, x = \frac{- 3 - 1}{2}

(x + 3)^{2} = 100

\frac{26 - 4 x}{x} < 9

496 - 624

8 x^{2} + 18 x - 5

\frac{- 1}{4} x + 8 \geq 12

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