What are the solutions to the equation 3(3x-9)^2+20=41 ?

The solutions to the equation 3(3x-9)^2+20=41 are x=(sqrt(7)+9)/3 ,x=(-sqrt(7)+9)/3

Find the zeros of 3(3x-9)^2+20=41

The zeros of 3(3x-9)^2+20=41 are x=(sqrt(7)+9)/3 ,x=(-sqrt(7)+9)/3

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3 (3 x - 9)^{2} + 20 = 41

x = \frac{7 + 9}{3}, x = \frac{- 7 + 9}{3}

Decimal Notation

x = 3.88191 \dots, x = 2.11808 \dots

Solution steps

3 (3 x - 9)^{2} + 20 = 41

Move

20

to the right side

3 (3 x - 9)^{2} = 21

Divide both sides by

3

(3 x - 9)^{2} = 7

For

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

the solutions are

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

Solve

3 x - 9 = 7 : x = \frac{7 + 9}{3}

Solve

3 x - 9 = - 7 : x = \frac{- 7 + 9}{3}

The solutions to the quadratic equation are:

x = \frac{7 + 9}{3}, x = \frac{- 7 + 9}{3}

- 2 x^{2} - 12 x = 0

2 p^{2} - p = 91

x^{2} + \frac{5}{2} x - 5 = 0

x, 2 x - x^{2} = 0.5

3 x^{2} - 9 a x^{2} - x + 3 a

What are the solutions to the equation 3(3x-9)^2+20=41 ?
The solutions to the equation 3(3x-9)^2+20=41 are x=(sqrt(7)+9)/3 ,x=(-sqrt(7)+9)/3
Find the zeros of 3(3x-9)^2+20=41
The zeros of 3(3x-9)^2+20=41 are x=(sqrt(7)+9)/3 ,x=(-sqrt(7)+9)/3