What are the solutions to the equation 3(3x-3)^2+11=41 ?

The solutions to the equation 3(3x-3)^2+11=41 are x=(sqrt(10)+3)/3 ,x=(-sqrt(10)+3)/3

Find the zeros of 3(3x-3)^2+11=41

The zeros of 3(3x-3)^2+11=41 are x=(sqrt(10)+3)/3 ,x=(-sqrt(10)+3)/3

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3 (3 x - 3)^{2} + 11 = 41

x = \frac{10 + 3}{3}, x = \frac{- 10 + 3}{3}

Decimal Notation

x = 2.05409 \dots, x = - 0.05409 \dots

Solution steps

3 (3 x - 3)^{2} + 11 = 41

Move

11

to the right side

3 (3 x - 3)^{2} = 30

Divide both sides by

3

(3 x - 3)^{2} = 10

For

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

the solutions are

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

Solve

3 x - 3 = 10 : x = \frac{10 + 3}{3}

Solve

3 x - 3 = - 10 : x = \frac{- 10 + 3}{3}

The solutions to the quadratic equation are:

x = \frac{10 + 3}{3}, x = \frac{- 10 + 3}{3}

- 11 s^{2} + 26 s + 61 = 0

(12 mn) (- m^{3} n^{- 2} p^{5}) (2 m)

3 (2 y - 3) = 7

x^{2} + x y - 5 x - 5 y

4 - x^{2} > x + 7

What are the solutions to the equation 3(3x-3)^2+11=41 ?
The solutions to the equation 3(3x-3)^2+11=41 are x=(sqrt(10)+3)/3 ,x=(-sqrt(10)+3)/3
Find the zeros of 3(3x-3)^2+11=41
The zeros of 3(3x-3)^2+11=41 are x=(sqrt(10)+3)/3 ,x=(-sqrt(10)+3)/3