What are the solutions to the equation 14a^2+4=2a ?

The solutions to the equation 14a^2+4=2a are a= 1/14+i(sqrt(55))/(14),a= 1/14-i(sqrt(55))/(14)

Find the zeros of 14a^2+4=2a

The zeros of 14a^2+4=2a are a= 1/14+i(sqrt(55))/(14),a= 1/14-i(sqrt(55))/(14)

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14 a^{2} + 4 = 2 a

a = \frac{1}{14} + i \frac{55}{14}, a = \frac{1}{14} - i \frac{55}{14}

Solution steps

14 a^{2} + 4 = 2 a

Move

2 a

to the left side

14 a^{2} + 4 - 2 a = 0

Write in the standard form

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

14 a^{2} - 2 a + 4 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 14 \cdot 4}{2 \cdot 14}

Simplify

(- 2)^{2} - 4 \cdot 14 \cdot 4 : 255 i

a_{1, 2} = \frac{- ( - 2 ) \pm 2 55 i}{2 \cdot 14}

Separate the solutions

a_{1} = \frac{- ( - 2 ) + 2 55 i}{2 \cdot 14}, a_{2} = \frac{- ( - 2 ) - 2 55 i}{2 \cdot 14}

a = \frac{- ( - 2 ) + 2 55 i}{2 \cdot 14} : \frac{1}{14} + i \frac{55}{14}

a = \frac{- ( - 2 ) - 2 55 i}{2 \cdot 14} : \frac{1}{14} - i \frac{55}{14}

The solutions to the quadratic equation are:

a = \frac{1}{14} + i \frac{55}{14}, a = \frac{1}{14} - i \frac{55}{14}

3 k^{2} + 6 k - 3

5 x - 3 = 2 x

\frac{x - 14}{- 4} = 3

42 v^{2} - 85 v + 42

x - \frac{x - 1}{3} - 1 = \frac{21 x - 22}{27}

What are the solutions to the equation 14a^2+4=2a ?
The solutions to the equation 14a^2+4=2a are a= 1/14+i(sqrt(55))/(14),a= 1/14-i(sqrt(55))/(14)
Find the zeros of 14a^2+4=2a
The zeros of 14a^2+4=2a are a= 1/14+i(sqrt(55))/(14),a= 1/14-i(sqrt(55))/(14)