What are the solutions to the equation 9x^2+6=2x ?

The solutions to the equation 9x^2+6=2x are x= 1/9+i(sqrt(53))/9 ,x= 1/9-i(sqrt(53))/9

Find the zeros of 9x^2+6=2x

The zeros of 9x^2+6=2x are x= 1/9+i(sqrt(53))/9 ,x= 1/9-i(sqrt(53))/9

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9 x^{2} + 6 = 2 x

x = \frac{1}{9} + i \frac{53}{9}, x = \frac{1}{9} - i \frac{53}{9}

Solution steps

9 x^{2} + 6 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 9 \cdot 6 : 253 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 53 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 53 i}{2 \cdot 9}, x_{2} = \frac{- ( - 2 ) - 2 53 i}{2 \cdot 9}

x = \frac{- ( - 2 ) + 2 53 i}{2 \cdot 9} : \frac{1}{9} + i \frac{53}{9}

x = \frac{- ( - 2 ) - 2 53 i}{2 \cdot 9} : \frac{1}{9} - i \frac{53}{9}

The solutions to the quadratic equation are:

x = \frac{1}{9} + i \frac{53}{9}, x = \frac{1}{9} - i \frac{53}{9}

2 x^{2} + x - 528 = 0

x^{2} + 10 x + 27 = 0

x^{2} + 4 x = 65

24 t^{2} - 24 = 0

x^{2} + 4 x = 56

What are the solutions to the equation 9x^2+6=2x ?
The solutions to the equation 9x^2+6=2x are x= 1/9+i(sqrt(53))/9 ,x= 1/9-i(sqrt(53))/9
Find the zeros of 9x^2+6=2x
The zeros of 9x^2+6=2x are x= 1/9+i(sqrt(53))/9 ,x= 1/9-i(sqrt(53))/9