What are the solutions to the equation 3(g-4)^2=27 ?

The solutions to the equation 3(g-4)^2=27 are g=7,g=1

Find the zeros of 3(g-4)^2=27

The zeros of 3(g-4)^2=27 are g=7,g=1

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3 (g - 4)^{2} = 27

g = 7, g = 1

Solution steps

3 (g - 4)^{2} = 27

Divide both sides by

3

(g - 4)^{2} = 9

For

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

the solutions are

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

Solve

g - 4 = 9 : g = 7

Solve

g - 4 = - 9 : g = 1

The solutions to the quadratic equation are:

g = 7, g = 1

70 x^{2} + 120 x = 440

2 x^{2} - 14 x + 10 = 26

x^{2} - 14 x + 35 = - 10

v, E = m g h + \frac{1}{2} m v^{2}

x, x (y^{2} + 3) = (y^{2} x^{2}) + 3

What are the solutions to the equation 3(g-4)^2=27 ?
The solutions to the equation 3(g-4)^2=27 are g=7,g=1
Find the zeros of 3(g-4)^2=27
The zeros of 3(g-4)^2=27 are g=7,g=1