What are the solutions to the equation 2x^2+9+3x=5x ?

The solutions to the equation 2x^2+9+3x=5x are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2

Find the zeros of 2x^2+9+3x=5x

The zeros of 2x^2+9+3x=5x are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2

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2 x^{2} + 9 + 3 x = 5 x

x = \frac{1}{2} + i \frac{17}{2}, x = \frac{1}{2} - i \frac{17}{2}

Solution steps

2 x^{2} + 9 + 3 x = 5 x

Move

5 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 9 : 217 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 17 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 17 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 17 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 17 i}{2 \cdot 2} : \frac{1}{2} + i \frac{17}{2}

x = \frac{- ( - 2 ) - 2 17 i}{2 \cdot 2} : \frac{1}{2} - i \frac{17}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{17}{2}, x = \frac{1}{2} - i \frac{17}{2}

x^{2} - 3 x + 20 = 0

x, x^{2} + y^{2} + z^{2} = 0

4 x^{2} - 16 x - 3

- 0.2 x^{2} + 8 x + 20 = - 2.4 x + 97.4

6 x - x^{2} = 2 x

What are the solutions to the equation 2x^2+9+3x=5x ?
The solutions to the equation 2x^2+9+3x=5x are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2
Find the zeros of 2x^2+9+3x=5x
The zeros of 2x^2+9+3x=5x are x= 1/2+i(sqrt(17))/2 ,x= 1/2-i(sqrt(17))/2