What are the solutions to the equation 9(4-x)^2=27 ?

The solutions to the equation 9(4-x)^2=27 are x=4-sqrt(3),x=4+sqrt(3)

Find the zeros of 9(4-x)^2=27

The zeros of 9(4-x)^2=27 are x=4-sqrt(3),x=4+sqrt(3)

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9 (4 - x)^{2} = 27

x = 4 - 3, x = 4 + 3

Decimal Notation

x = 2.26794 \dots, x = 5.73205 \dots

Solution steps

9 (4 - x)^{2} = 27

Divide both sides by

9

(4 - x)^{2} = 3

For

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

the solutions are

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

Solve

4 - x = 3 : x = 4 - 3

Solve

4 - x = - 3 : x = 4 + 3

The solutions to the quadratic equation are:

x = 4 - 3, x = 4 + 3

x, y^{2} + x^{2} = c

64 - x^{2} = - 36

x (x + 5) - 2 x = 0

4 (x^{2} - 2) - 3 (x^{2} - 1) = 20

x, 36 y^{2} + x^{2} + 36 z = 9

What are the solutions to the equation 9(4-x)^2=27 ?
The solutions to the equation 9(4-x)^2=27 are x=4-sqrt(3),x=4+sqrt(3)
Find the zeros of 9(4-x)^2=27
The zeros of 9(4-x)^2=27 are x=4-sqrt(3),x=4+sqrt(3)