What are the solutions to the equation 2(y+4)^2=22 ?

The solutions to the equation 2(y+4)^2=22 are y=sqrt(11)-4,y=-sqrt(11)-4

Find the zeros of 2(y+4)^2=22

The zeros of 2(y+4)^2=22 are y=sqrt(11)-4,y=-sqrt(11)-4

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2 (y + 4)^{2} = 22

y = 11 - 4, y = - 11 - 4

Decimal Notation

y = - 0.68337 \dots, y = - 7.31662 \dots

Solution steps

2 (y + 4)^{2} = 22

Divide both sides by

2

(y + 4)^{2} = 11

For

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

the solutions are

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

Solve

y + 4 = 11 : y = 11 - 4

Solve

y + 4 = - 11 : y = - 11 - 4

The solutions to the quadratic equation are:

y = 11 - 4, y = - 11 - 4

1 0^{1}

8 (1) - 2 (- 5) \geq - 12

7 b + 9 = 6 b + 14

- 7 (7 + 6 r) > 245

- 9 x - 30 x^{2}

What are the solutions to the equation 2(y+4)^2=22 ?
The solutions to the equation 2(y+4)^2=22 are y=sqrt(11)-4,y=-sqrt(11)-4
Find the zeros of 2(y+4)^2=22
The zeros of 2(y+4)^2=22 are y=sqrt(11)-4,y=-sqrt(11)-4