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

The solutions to the equation 3(x^2+x)=2(x+2) are x=1,x=-4/3

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

The zeros of 3(x^2+x)=2(x+2) are x=1,x=-4/3

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

x = 1, x = - \frac{4}{3}

Decimal Notation

x = 1, x = - 1.33333 \dots

Solution steps

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

Expand

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

Expand

2 (x + 2) : 2 x + 4

3 x^{2} + 3 x = 2 x + 4

Move

4

to the left side

3 x^{2} + 3 x - 4 = 2 x

Move

2 x

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 3 (- 4) = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 \cdot 3}, x_{2} = \frac{- 1 - 7}{2 \cdot 3}

x = \frac{- 1 + 7}{2 \cdot 3} : 1

x = \frac{- 1 - 7}{2 \cdot 3} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = 1, x = - \frac{4}{3}

\frac{x ^{2} - 5}{x} < - 4

3 w^{3} + 192

\frac{2}{5} x^{2} y^{2} + \frac{1}{5} x^{2} y^{2}

3 z^{2} + 23 z + 30

- x^{3} + a^{3}

What are the solutions to the equation 3(x^2+x)=2(x+2) ?
The solutions to the equation 3(x^2+x)=2(x+2) are x=1,x=-4/3
Find the zeros of 3(x^2+x)=2(x+2)
The zeros of 3(x^2+x)=2(x+2) are x=1,x=-4/3