What are the solutions to the equation (2y+20)^2=27 ?

The solutions to the equation (2y+20)^2=27 are y=(3sqrt(3)-20)/2 ,y=(-3sqrt(3)-20)/2

Find the zeros of (2y+20)^2=27

The zeros of (2y+20)^2=27 are y=(3sqrt(3)-20)/2 ,y=(-3sqrt(3)-20)/2

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(2 y + 20)^{2} = 27

y = \frac{3 3 - 20}{2}, y = \frac{- 3 3 - 20}{2}

Decimal Notation

y = - 7.40192 \dots, y = - 12.59807 \dots

Solution steps

(2 y + 20)^{2} = 27

For

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

the solutions are

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

Solve

2 y + 20 = 27 : y = \frac{3 3 - 20}{2}

Solve

2 y + 20 = - 27 : y = \frac{- 3 3 - 20}{2}

The solutions to the quadratic equation are:

y = \frac{3 3 - 20}{2}, y = \frac{- 3 3 - 20}{2}

- x^{2} + 6 x + 1

x^{2} - 144 x = 0

3 x^{2} + 12 x + 28 = 0

(x + 2) (2 x - 5) = 0

2 (x - 6)^{2} + 5 = 23

What are the solutions to the equation (2y+20)^2=27 ?
The solutions to the equation (2y+20)^2=27 are y=(3sqrt(3)-20)/2 ,y=(-3sqrt(3)-20)/2
Find the zeros of (2y+20)^2=27
The zeros of (2y+20)^2=27 are y=(3sqrt(3)-20)/2 ,y=(-3sqrt(3)-20)/2