What are the solutions to the equation (2k+3)^2=27 ?

The solutions to the equation (2k+3)^2=27 are k=(3sqrt(3)-3)/2 ,k=(-3sqrt(3)-3)/2

Find the zeros of (2k+3)^2=27

The zeros of (2k+3)^2=27 are k=(3sqrt(3)-3)/2 ,k=(-3sqrt(3)-3)/2

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(2 k + 3)^{2} = 27

k = \frac{3 3 - 3}{2}, k = \frac{- 3 3 - 3}{2}

Decimal Notation

k = 1.09807 \dots, k = - 4.09807 \dots

Solution steps

(2 k + 3)^{2} = 27

For

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

the solutions are

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

Solve

2 k + 3 = 27 : k = \frac{3 3 - 3}{2}

Solve

2 k + 3 = - 27 : k = \frac{- 3 3 - 3}{2}

The solutions to the quadratic equation are:

k = \frac{3 3 - 3}{2}, k = \frac{- 3 3 - 3}{2}

1 6^{2} + x^{2} = 2 0^{2}

∣ 2 x ∣ = 10

4 (0) + 1

- ∣ 2 x + 1 ∣ = - 3

x^{2} + 13 x + 9 = 0

What are the solutions to the equation (2k+3)^2=27 ?
The solutions to the equation (2k+3)^2=27 are k=(3sqrt(3)-3)/2 ,k=(-3sqrt(3)-3)/2
Find the zeros of (2k+3)^2=27
The zeros of (2k+3)^2=27 are k=(3sqrt(3)-3)/2 ,k=(-3sqrt(3)-3)/2