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

The solutions to the equation 3x(x+1)=2x-7 are x=-1/6+i(sqrt(83))/6 ,x=-1/6-i(sqrt(83))/6

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

The zeros of 3x(x+1)=2x-7 are x=-1/6+i(sqrt(83))/6 ,x=-1/6-i(sqrt(83))/6

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3 x (x + 1) = 2 x - 7

x = - \frac{1}{6} + i \frac{83}{6}, x = - \frac{1}{6} - i \frac{83}{6}

Solution steps

3 x (x + 1) = 2 x - 7

Expand

3 x (x + 1) : 3 x^{2} + 3 x

3 x^{2} + 3 x = 2 x - 7

Move

7

to the left side

3 x^{2} + 3 x + 7 = 2 x

Move

2 x

to the left side

3 x^{2} + x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 \cdot 7}{2 \cdot 3}

Simplify

1^{2} - 4 \cdot 3 \cdot 7 : 83 i

x_{1, 2} = \frac{- 1 \pm 83 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 83 i}{2 \cdot 3}, x_{2} = \frac{- 1 - 83 i}{2 \cdot 3}

x = \frac{- 1 + 83 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{83}{6}

x = \frac{- 1 - 83 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{83}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{6} + i \frac{83}{6}, x = - \frac{1}{6} - i \frac{83}{6}

x^{2} + 8 x + 15 = 0

(m - 3) (3 m + 4) = 5 (m + 1) + 8

- 2 294 m^{16} n^{7}

\frac{3 x y ^{2}}{5 x y ^{3}}

x^{3} + x^{2} - 14 x - 24

What are the solutions to the equation 3x(x+1)=2x-7 ?
The solutions to the equation 3x(x+1)=2x-7 are x=-1/6+i(sqrt(83))/6 ,x=-1/6-i(sqrt(83))/6
Find the zeros of 3x(x+1)=2x-7
The zeros of 3x(x+1)=2x-7 are x=-1/6+i(sqrt(83))/6 ,x=-1/6-i(sqrt(83))/6