What are the solutions to the equation (4y+12)^2=32 ?

The solutions to the equation (4y+12)^2=32 are y=sqrt(2)-3,y=-3-sqrt(2)

Find the zeros of (4y+12)^2=32

The zeros of (4y+12)^2=32 are y=sqrt(2)-3,y=-3-sqrt(2)

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(4 y + 12)^{2} = 32

y = 2 - 3, y = - 3 - 2

Decimal Notation

y = - 1.58578 \dots, y = - 4.41421 \dots

Solution steps

(4 y + 12)^{2} = 32

For

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

the solutions are

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

Solve

4 y + 12 = 32 : y = 2 - 3

Solve

4 y + 12 = - 32 : y = - 3 - 2

The solutions to the quadratic equation are:

y = 2 - 3, y = - 3 - 2

6 x^{2} - 6 x + 9 = 0

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

- x^{2} = - 25

5 x^{2} = 37 x + 72

x, 15 x^{2} = 38 x - 24

What are the solutions to the equation (4y+12)^2=32 ?
The solutions to the equation (4y+12)^2=32 are y=sqrt(2)-3,y=-3-sqrt(2)
Find the zeros of (4y+12)^2=32
The zeros of (4y+12)^2=32 are y=sqrt(2)-3,y=-3-sqrt(2)