What are the solutions to the equation (4x+1)^2=5 ?

The solutions to the equation (4x+1)^2=5 are x=(sqrt(5)-1)/4 ,x=(-sqrt(5)-1)/4

Find the zeros of (4x+1)^2=5

The zeros of (4x+1)^2=5 are x=(sqrt(5)-1)/4 ,x=(-sqrt(5)-1)/4

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(4 x + 1)^{2} = 5

x = \frac{5 - 1}{4}, x = \frac{- 5 - 1}{4}

Decimal Notation

x = 0.30901 \dots, x = - 0.80901 \dots

Solution steps

(4 x + 1)^{2} = 5

For

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

the solutions are

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

Solve

4 x + 1 = 5 : x = \frac{5 - 1}{4}

Solve

4 x + 1 = - 5 : x = \frac{- 5 - 1}{4}

The solutions to the quadratic equation are:

x = \frac{5 - 1}{4}, x = \frac{- 5 - 1}{4}

\frac{1}{3} x + \frac{1}{2} = - 4

- 9 - x = - 5

(x - 5) \cdot (x - 4) \leq 6

5 \cdot 6

x^{2} + 0 x + 1 = 0

What are the solutions to the equation (4x+1)^2=5 ?
The solutions to the equation (4x+1)^2=5 are x=(sqrt(5)-1)/4 ,x=(-sqrt(5)-1)/4
Find the zeros of (4x+1)^2=5
The zeros of (4x+1)^2=5 are x=(sqrt(5)-1)/4 ,x=(-sqrt(5)-1)/4