What are the solutions to the equation (2x+1)^2=12 ?

The solutions to the equation (2x+1)^2=12 are x=(2sqrt(3)-1)/2 ,x=(-2sqrt(3)-1)/2

Find the zeros of (2x+1)^2=12

The zeros of (2x+1)^2=12 are x=(2sqrt(3)-1)/2 ,x=(-2sqrt(3)-1)/2

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(2 x + 1)^{2} = 12

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

Decimal Notation

x = 1.23205 \dots, x = - 2.23205 \dots

Solution steps

(2 x + 1)^{2} = 12

For

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

the solutions are

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

Solve

2 x + 1 = 12 : x = \frac{2 3 - 1}{2}

Solve

2 x + 1 = - 12 : x = \frac{- 2 3 - 1}{2}

The solutions to the quadratic equation are:

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

4 v^{2} - 64

(\frac{3 x}{y})^{- 3}

∣ 3 x - 4 ∣ \geq 2

6 (2 m - 3) - 3 m = 2 m - 18 + m

(x - 2)^{2} < (x - 4) (x + 4)

What are the solutions to the equation (2x+1)^2=12 ?
The solutions to the equation (2x+1)^2=12 are x=(2sqrt(3)-1)/2 ,x=(-2sqrt(3)-1)/2
Find the zeros of (2x+1)^2=12
The zeros of (2x+1)^2=12 are x=(2sqrt(3)-1)/2 ,x=(-2sqrt(3)-1)/2