What are the solutions to the equation (4x+1)^2=20 ?

The solutions to the equation (4x+1)^2=20 are x=(2sqrt(5)-1)/4 ,x=(-2sqrt(5)-1)/4

Find the zeros of (4x+1)^2=20

The zeros of (4x+1)^2=20 are x=(2sqrt(5)-1)/4 ,x=(-2sqrt(5)-1)/4

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(4 x + 1)^{2} = 20

x = \frac{2 5 - 1}{4}, x = \frac{- 2 5 - 1}{4}

Decimal Notation

x = 0.86803 \dots, x = - 1.36803 \dots

Solution steps

(4 x + 1)^{2} = 20

For

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

the solutions are

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

Solve

4 x + 1 = 20 : x = \frac{2 5 - 1}{4}

Solve

4 x + 1 = - 20 : x = \frac{- 2 5 - 1}{4}

The solutions to the quadratic equation are:

x = \frac{2 5 - 1}{4}, x = \frac{- 2 5 - 1}{4}

∣ 3 x + 6 ∣ = 15

(- \frac{3}{2})^{2} + 8 (- \frac{3}{2})

\frac{4 π}{3} - 2 π

64 x^{3} - 27

6 x = 54

What are the solutions to the equation (4x+1)^2=20 ?
The solutions to the equation (4x+1)^2=20 are x=(2sqrt(5)-1)/4 ,x=(-2sqrt(5)-1)/4
Find the zeros of (4x+1)^2=20
The zeros of (4x+1)^2=20 are x=(2sqrt(5)-1)/4 ,x=(-2sqrt(5)-1)/4