What are the solutions to the equation 10x^2+9=x ?

The solutions to the equation 10x^2+9=x are x= 1/20+i(sqrt(359))/(20),x= 1/20-i(sqrt(359))/(20)

Find the zeros of 10x^2+9=x

The zeros of 10x^2+9=x are x= 1/20+i(sqrt(359))/(20),x= 1/20-i(sqrt(359))/(20)

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10 x^{2} + 9 = x

x = \frac{1}{20} + i \frac{359}{20}, x = \frac{1}{20} - i \frac{359}{20}

Solution steps

10 x^{2} + 9 = x

Move

x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 10 \cdot 9}{2 \cdot 10}

Simplify

(- 1)^{2} - 4 \cdot 10 \cdot 9 : 359 i

x_{1, 2} = \frac{- ( - 1 ) \pm 359 i}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 359 i}{2 \cdot 10}, x_{2} = \frac{- ( - 1 ) - 359 i}{2 \cdot 10}

x = \frac{- ( - 1 ) + 359 i}{2 \cdot 10} : \frac{1}{20} + i \frac{359}{20}

x = \frac{- ( - 1 ) - 359 i}{2 \cdot 10} : \frac{1}{20} - i \frac{359}{20}

The solutions to the quadratic equation are:

x = \frac{1}{20} + i \frac{359}{20}, x = \frac{1}{20} - i \frac{359}{20}

8^{2}

\frac{2 x ^{- 2} y ^{- 2}}{16 x ^{- 12} y ^{2}}

55 x \geq 505 x + 3150

5 x + 3 = 10

4 + 9

What are the solutions to the equation 10x^2+9=x ?
The solutions to the equation 10x^2+9=x are x= 1/20+i(sqrt(359))/(20),x= 1/20-i(sqrt(359))/(20)
Find the zeros of 10x^2+9=x
The zeros of 10x^2+9=x are x= 1/20+i(sqrt(359))/(20),x= 1/20-i(sqrt(359))/(20)