What are the solutions to the equation (2x+1)^2=31 ?

The solutions to the equation (2x+1)^2=31 are x=(sqrt(31)-1)/2 ,x=(-sqrt(31)-1)/2

Find the zeros of (2x+1)^2=31

The zeros of (2x+1)^2=31 are x=(sqrt(31)-1)/2 ,x=(-sqrt(31)-1)/2

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(2 x + 1)^{2} = 31

x = \frac{31 - 1}{2}, x = \frac{- 31 - 1}{2}

Decimal Notation

x = 2.28388 \dots, x = - 3.28388 \dots

Solution steps

(2 x + 1)^{2} = 31

For

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

the solutions are

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

Solve

2 x + 1 = 31 : x = \frac{31 - 1}{2}

Solve

2 x + 1 = - 31 : x = \frac{- 31 - 1}{2}

The solutions to the quadratic equation are:

x = \frac{31 - 1}{2}, x = \frac{- 31 - 1}{2}

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

(m - 1) x^{2} - 3 m x + m + 1 = 0

42 x^{2} + 11 x - 20 = 0

x, (x - a)^{2} + y^{2} = a^{2}

16.1 x^{2} + 18 x = 50

What are the solutions to the equation (2x+1)^2=31 ?
The solutions to the equation (2x+1)^2=31 are x=(sqrt(31)-1)/2 ,x=(-sqrt(31)-1)/2
Find the zeros of (2x+1)^2=31
The zeros of (2x+1)^2=31 are x=(sqrt(31)-1)/2 ,x=(-sqrt(31)-1)/2