What are the solutions to the equation 3(2x-4)^2=36 ?

The solutions to the equation 3(2x-4)^2=36 are x=sqrt(3)+2,x=-sqrt(3)+2

Find the zeros of 3(2x-4)^2=36

The zeros of 3(2x-4)^2=36 are x=sqrt(3)+2,x=-sqrt(3)+2

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3 (2 x - 4)^{2} = 36

x = 3 + 2, x = - 3 + 2

Decimal Notation

x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

3 (2 x - 4)^{2} = 36

Divide both sides by

3

(2 x - 4)^{2} = 12

For

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

the solutions are

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

Solve

2 x - 4 = 12 : x = 3 + 2

Solve

2 x - 4 = - 12 : x = - 3 + 2

The solutions to the quadratic equation are:

x = 3 + 2, x = - 3 + 2

(y - 2) (y + 5) = (2 y - 1) (y + 1) + 7

x^{2} + 10 x + 24 = 0

x^{2} + 13 x

t^{2} - 8 t + 16 = 0

x^{2} + 4 x = - 9

What are the solutions to the equation 3(2x-4)^2=36 ?
The solutions to the equation 3(2x-4)^2=36 are x=sqrt(3)+2,x=-sqrt(3)+2
Find the zeros of 3(2x-4)^2=36
The zeros of 3(2x-4)^2=36 are x=sqrt(3)+2,x=-sqrt(3)+2