What are the solutions to the equation (3x+2)^2-5(3x+2)=24 ?

The solutions to the equation (3x+2)^2-5(3x+2)=24 are x=2,x=-5/3

Find the zeros of (3x+2)^2-5(3x+2)=24

The zeros of (3x+2)^2-5(3x+2)=24 are x=2,x=-5/3

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(3 x + 2)^{2} - 5 (3 x + 2) = 24

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

Decimal Notation

x = 2, x = - 1.66666 \dots

Solution steps

(3 x + 2)^{2} - 5 (3 x + 2) = 24

Expand

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

9 x^{2} - 3 x - 6 = 24

Move

24

to the left side

9 x^{2} - 3 x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 30 )}{2 \cdot 9}

(- 3)^{2} - 4 \cdot 9 (- 30) = 33

x_{1, 2} = \frac{- ( - 3 ) \pm 33}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 33}{2 \cdot 9}, x_{2} = \frac{- ( - 3 ) - 33}{2 \cdot 9}

x = \frac{- ( - 3 ) + 33}{2 \cdot 9} : 2

x = \frac{- ( - 3 ) - 33}{2 \cdot 9} : - \frac{5}{3}

The solutions to the quadratic equation are:

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

3 x^{2} + 2.25 x - 48.18 = 0

3 t^{2} - 16 t + 24 = 0

- \frac{121 K - 241920}{756} > 0

\frac{3}{3} - \frac{1}{3}

5 x + 54 \geq 109

What are the solutions to the equation (3x+2)^2-5(3x+2)=24 ?
The solutions to the equation (3x+2)^2-5(3x+2)=24 are x=2,x=-5/3
Find the zeros of (3x+2)^2-5(3x+2)=24
The zeros of (3x+2)^2-5(3x+2)=24 are x=2,x=-5/3