What are the solutions to the equation 3(3x+1)^2-12=-9 ?

The solutions to the equation 3(3x+1)^2-12=-9 are x=0,x=-2/3

Find the zeros of 3(3x+1)^2-12=-9

The zeros of 3(3x+1)^2-12=-9 are x=0,x=-2/3

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3 (3 x + 1)^{2} - 12 = - 9

x = 0, x = - \frac{2}{3}

Decimal Notation

x = 0, x = - 0.66666 \dots

Solution steps

3 (3 x + 1)^{2} - 12 = - 9

Move

12

to the right side

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

Divide both sides by

3

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

For

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

the solutions are

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

Solve

3 x + 1 = 1 : x = 0

Solve

3 x + 1 = - 1 : x = - \frac{2}{3}

The solutions to the quadratic equation are:

x = 0, x = - \frac{2}{3}

3 x + 2 > - 3

\frac{6}{3 - 5}

4 x - 6 = 2 (x - 1) + x

\frac{12 !}{7 ! ( 12 - 7 ) !}

a^{5} + 1

What are the solutions to the equation 3(3x+1)^2-12=-9 ?
The solutions to the equation 3(3x+1)^2-12=-9 are x=0,x=-2/3
Find the zeros of 3(3x+1)^2-12=-9
The zeros of 3(3x+1)^2-12=-9 are x=0,x=-2/3