What are the solutions to the equation 3(x+1)^2=12 ?

The solutions to the equation 3(x+1)^2=12 are x=1,x=-3

Find the zeros of 3(x+1)^2=12

The zeros of 3(x+1)^2=12 are x=1,x=-3

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3 (x + 1)^{2} = 12

x = 1, x = - 3

Solution steps

3 (x + 1)^{2} = 12

Divide both sides by

3

(x + 1)^{2} = 4

For

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

the solutions are

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

Solve

x + 1 = 4 : x = 1

Solve

x + 1 = - 4 : x = - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

x \cdot (x - 1) = 72

25 y^{2} - 15 y + 9 = 0

x^{2} + 3 x - 6 = 0

x, x^{2} + y^{2} - 2 y + 6 = 0

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

What are the solutions to the equation 3(x+1)^2=12 ?
The solutions to the equation 3(x+1)^2=12 are x=1,x=-3
Find the zeros of 3(x+1)^2=12
The zeros of 3(x+1)^2=12 are x=1,x=-3