What are the solutions to the equation 3(2x+1)^2+2=7 ?

The solutions to the equation 3(2x+1)^2+2=7 are x=(sqrt(5/3)-1)/2 ,x=(-sqrt(5/3)-1)/2

Find the zeros of 3(2x+1)^2+2=7

The zeros of 3(2x+1)^2+2=7 are x=(sqrt(5/3)-1)/2 ,x=(-sqrt(5/3)-1)/2

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3 (2 x + 1)^{2} + 2 = 7

x = \frac{\frac{5}{3} - 1}{2}, x = \frac{- \frac{5}{3} - 1}{2}

Decimal Notation

x = 0.14549 \dots, x = - 1.14549 \dots

Solution steps

3 (2 x + 1)^{2} + 2 = 7

Move

2

to the right side

3 (2 x + 1)^{2} = 5

Divide both sides by

3

(2 x + 1)^{2} = \frac{5}{3}

For

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

the solutions are

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

Solve

2 x + 1 = \frac{5}{3} : x = \frac{\frac{5}{3} - 1}{2}

Solve

2 x + 1 = - \frac{5}{3} : x = \frac{- \frac{5}{3} - 1}{2}

The solutions to the quadratic equation are:

x = \frac{\frac{5}{3} - 1}{2}, x = \frac{- \frac{5}{3} - 1}{2}

w^{2} + 50 = - 15 w

4 x^{2} + 7 x - 7 = 0

3 x^{2} - 5 x + 1

(6 x - 3) (2 x + 4) = 3

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

What are the solutions to the equation 3(2x+1)^2+2=7 ?
The solutions to the equation 3(2x+1)^2+2=7 are x=(sqrt(5/3)-1)/2 ,x=(-sqrt(5/3)-1)/2
Find the zeros of 3(2x+1)^2+2=7
The zeros of 3(2x+1)^2+2=7 are x=(sqrt(5/3)-1)/2 ,x=(-sqrt(5/3)-1)/2