What are the solutions to the equation 10v^2-1v+9=0 ?

The solutions to the equation 10v^2-1v+9=0 are v= 1/20+i(sqrt(359))/(20),v= 1/20-i(sqrt(359))/(20)

Find the zeros of 10v^2-1v+9=0

The zeros of 10v^2-1v+9=0 are v= 1/20+i(sqrt(359))/(20),v= 1/20-i(sqrt(359))/(20)

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10 v^{2} - 1 v + 9 = 0

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

Solution steps

10 v^{2} - 1 \cdot v + 9 = 0

Refine

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

Solve with the quadratic formula

v_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 x - 6 - 2 < 0

\frac{x}{4} = 16

(2 x - 2)^{2} + 1 = 6 - (x - 1)^{2}

x^{2} > 6 x - 9

4 a - 7 < 2 a + 1

What are the solutions to the equation 10v^2-1v+9=0 ?
The solutions to the equation 10v^2-1v+9=0 are v= 1/20+i(sqrt(359))/(20),v= 1/20-i(sqrt(359))/(20)
Find the zeros of 10v^2-1v+9=0
The zeros of 10v^2-1v+9=0 are v= 1/20+i(sqrt(359))/(20),v= 1/20-i(sqrt(359))/(20)