What are the solutions to the equation 2(3x+1)^2-32=0 ?

The solutions to the equation 2(3x+1)^2-32=0 are x=1,x=-5/3

Find the zeros of 2(3x+1)^2-32=0

The zeros of 2(3x+1)^2-32=0 are x=1,x=-5/3

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2 (3 x + 1)^{2} - 32 = 0

x = 1, x = - \frac{5}{3}

Decimal Notation

x = 1, x = - 1.66666 \dots

Solution steps

2 (3 x + 1)^{2} - 32 = 0

Move

32

to the right side

2 (3 x + 1)^{2} = 32

Divide both sides by

2

(3 x + 1)^{2} = 16

For

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

the solutions are

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

Solve

3 x + 1 = 16 : x = 1

Solve

3 x + 1 = - 16 : x = - \frac{5}{3}

The solutions to the quadratic equation are:

x = 1, x = - \frac{5}{3}

2 w^{2} - 7 w - 4 = 0

(x + 3)^{2} + 4 (x + 2)^{2} = 2 5^{2}

5 = 3 x^{2}

x, x^{2} - 10 x + 20 = 0

(3 - 2 x)^{2} = 7

What are the solutions to the equation 2(3x+1)^2-32=0 ?
The solutions to the equation 2(3x+1)^2-32=0 are x=1,x=-5/3
Find the zeros of 2(3x+1)^2-32=0
The zeros of 2(3x+1)^2-32=0 are x=1,x=-5/3