What are the solutions to the equation 1/3 (x+3)^2-5=4 ?

The solutions to the equation 1/3 (x+3)^2-5=4 are x=3sqrt(3)-3,x=-3sqrt(3)-3

Find the zeros of 1/3 (x+3)^2-5=4

The zeros of 1/3 (x+3)^2-5=4 are x=3sqrt(3)-3,x=-3sqrt(3)-3

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\frac{1}{3} (x + 3)^{2} - 5 = 4

x = 33 - 3, x = - 33 - 3

Solution steps

\frac{1}{3} (x + 3)^{2} - 5 = 4

Move

5

to the right side

\frac{1}{3} (x + 3)^{2} = 9

Multiply both sides by

3

(x + 3)^{2} = 27

For

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

the solutions are

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

Solve

x + 3 = 27 : x = 33 - 3

Solve

x + 3 = - 27 : x = - 33 - 3

The solutions to the quadratic equation are:

x = 33 - 3, x = - 33 - 3

x - 4 = 7

5 x^{2} + 9 = 4

- x^{2} + 11 = 0

\frac{4 + i}{2 - 5 i}

\frac{18 x ^{7} y ^{8}}{30 x ^{6} y ^{6}}

What are the solutions to the equation 1/3 (x+3)^2-5=4 ?
The solutions to the equation 1/3 (x+3)^2-5=4 are x=3sqrt(3)-3,x=-3sqrt(3)-3
Find the zeros of 1/3 (x+3)^2-5=4
The zeros of 1/3 (x+3)^2-5=4 are x=3sqrt(3)-3,x=-3sqrt(3)-3