What are the solutions to the equation (3x+3)^2-42=-39 ?

The solutions to the equation (3x+3)^2-42=-39 are x=(1-sqrt(3))/(sqrt(3)),x=-(1+sqrt(3))/(sqrt(3))

Find the zeros of (3x+3)^2-42=-39

The zeros of (3x+3)^2-42=-39 are x=(1-sqrt(3))/(sqrt(3)),x=-(1+sqrt(3))/(sqrt(3))

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(3 x + 3)^{2} - 42 = - 39

x = \frac{1 - 3}{3}, x = - \frac{1 + 3}{3}

Decimal Notation

x = - 0.42264 \dots, x = - 1.57735 \dots

Solution steps

(3 x + 3)^{2} - 42 = - 39

Move

42

to the right side

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

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

x = \frac{1 - 3}{3}, x = - \frac{1 + 3}{3}

4 x^{2} - 8 x

x, U = \frac{1}{2} k x^{2}

12 x^{2} - 10 x + 1 = 0

x^{2} - 29 x = 0

3 x^{2} - 12 x = - 27

What are the solutions to the equation (3x+3)^2-42=-39 ?
The solutions to the equation (3x+3)^2-42=-39 are x=(1-sqrt(3))/(sqrt(3)),x=-(1+sqrt(3))/(sqrt(3))
Find the zeros of (3x+3)^2-42=-39
The zeros of (3x+3)^2-42=-39 are x=(1-sqrt(3))/(sqrt(3)),x=-(1+sqrt(3))/(sqrt(3))