What are the solutions to the equation 2x+3x^2=6x-5 ?

The solutions to the equation 2x+3x^2=6x-5 are x= 2/3+i(sqrt(11))/3 ,x= 2/3-i(sqrt(11))/3

Find the zeros of 2x+3x^2=6x-5

The zeros of 2x+3x^2=6x-5 are x= 2/3+i(sqrt(11))/3 ,x= 2/3-i(sqrt(11))/3

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2 x + 3 x^{2} = 6 x - 5

x = \frac{2}{3} + i \frac{11}{3}, x = \frac{2}{3} - i \frac{11}{3}

Solution steps

2 x + 3 x^{2} = 6 x - 5

Move

5

to the left side

2 x + 3 x^{2} + 5 = 6 x

Move

6 x

to the left side

3 x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 5 : 211 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 11 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 11 i}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 11 i}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 11 i}{2 \cdot 3} : \frac{2}{3} + i \frac{11}{3}

x = \frac{- ( - 4 ) - 2 11 i}{2 \cdot 3} : \frac{2}{3} - i \frac{11}{3}

The solutions to the quadratic equation are:

x = \frac{2}{3} + i \frac{11}{3}, x = \frac{2}{3} - i \frac{11}{3}

12 x^{2} - 780 x + 9000 = 0

\frac{4}{3} x^{2} - 4 x + \frac{10}{3} = 0

3 x^{2} - 9 x + 16 = 178

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

x^{2} - 22 x + 96 = 0

What are the solutions to the equation 2x+3x^2=6x-5 ?
The solutions to the equation 2x+3x^2=6x-5 are x= 2/3+i(sqrt(11))/3 ,x= 2/3-i(sqrt(11))/3
Find the zeros of 2x+3x^2=6x-5
The zeros of 2x+3x^2=6x-5 are x= 2/3+i(sqrt(11))/3 ,x= 2/3-i(sqrt(11))/3