What are the solutions to the equation (3x+2)^2-5(3x+2)=0 ?

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

Find the zeros of (3x+2)^2-5(3x+2)=0

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

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(3 x + 2)^{2} - 5 (3 x + 2) = 0

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

Solution steps

(3 x + 2)^{2} - 5 (3 x + 2) = 0

Expand

(3 x + 2)^{2} - 5 (3 x + 2) : 9 x^{2} - 3 x - 6

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 6 )}{2 \cdot 9}

(- 3)^{2} - 4 \cdot 9 (- 6) = 15

x_{1, 2} = \frac{- ( - 3 ) \pm 15}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 15}{2 \cdot 9}, x_{2} = \frac{- ( - 3 ) - 15}{2 \cdot 9}

x = \frac{- ( - 3 ) + 15}{2 \cdot 9} : 1

x = \frac{- ( - 3 ) - 15}{2 \cdot 9} : - \frac{2}{3}

The solutions to the quadratic equation are:

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

b^{2} + 3 b - 10 = 0

- 16 x^{2} + 32 x = 0

37 n^{2} + 42 n - 20 = - 12 n^{2} + 7

x (x + 1) = 4 - (x + 2)^{2}

3 x^{2} + 24 x + 32 = 0

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