What are the solutions to the equation 6x^2-3x+0.9=0 ?

The solutions to the equation 6x^2-3x+0.9=0 are x= 1/4+i(sqrt(35))/(20),x= 1/4-i(sqrt(35))/(20)

Find the zeros of 6x^2-3x+0.9=0

The zeros of 6x^2-3x+0.9=0 are x= 1/4+i(sqrt(35))/(20),x= 1/4-i(sqrt(35))/(20)

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6 x^{2} - 3 x + 0.9 = 0

x = \frac{1}{4} + i \frac{35}{20}, x = \frac{1}{4} - i \frac{35}{20}

Solution steps

6 x^{2} - 3 x + 0.9 = 0

Multiply both sides by

10

60 x^{2} - 30 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 60 \cdot 9}{2 \cdot 60}

Simplify

(- 30)^{2} - 4 \cdot 60 \cdot 9 : 635 i

x_{1, 2} = \frac{- ( - 30 ) \pm 6 35 i}{2 \cdot 60}

Separate the solutions

x_{1} = \frac{- ( - 30 ) + 6 35 i}{2 \cdot 60}, x_{2} = \frac{- ( - 30 ) - 6 35 i}{2 \cdot 60}

x = \frac{- ( - 30 ) + 6 35 i}{2 \cdot 60} : \frac{1}{4} + i \frac{35}{20}

x = \frac{- ( - 30 ) - 6 35 i}{2 \cdot 60} : \frac{1}{4} - i \frac{35}{20}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{35}{20}, x = \frac{1}{4} - i \frac{35}{20}

x, k x^{2} + 4 k x + 3 = x^{2}

x (2 x + 3) + 4 = x (x + 9)

1^{2} + y^{2} = 1 + 19 y

12 m^{2} + m - 10 = - 4

x^{2} = 14 x - 48

What are the solutions to the equation 6x^2-3x+0.9=0 ?
The solutions to the equation 6x^2-3x+0.9=0 are x= 1/4+i(sqrt(35))/(20),x= 1/4-i(sqrt(35))/(20)
Find the zeros of 6x^2-3x+0.9=0
The zeros of 6x^2-3x+0.9=0 are x= 1/4+i(sqrt(35))/(20),x= 1/4-i(sqrt(35))/(20)