What are the solutions to the equation 20x^2+1=x ?

The solutions to the equation 20x^2+1=x are x= 1/40+i(sqrt(79))/(40),x= 1/40-i(sqrt(79))/(40)

Find the zeros of 20x^2+1=x

The zeros of 20x^2+1=x are x= 1/40+i(sqrt(79))/(40),x= 1/40-i(sqrt(79))/(40)

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20 x^{2} + 1 = x

x = \frac{1}{40} + i \frac{79}{40}, x = \frac{1}{40} - i \frac{79}{40}

Solution steps

20 x^{2} + 1 = x

Move

x

to the left side

20 x^{2} + 1 - x = 0

Write in the standard form

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

20 x^{2} - x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 20 \cdot 1}{2 \cdot 20}

Simplify

(- 1)^{2} - 4 \cdot 20 \cdot 1 : 79 i

x_{1, 2} = \frac{- ( - 1 ) \pm 79 i}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 79 i}{2 \cdot 20}, x_{2} = \frac{- ( - 1 ) - 79 i}{2 \cdot 20}

x = \frac{- ( - 1 ) + 79 i}{2 \cdot 20} : \frac{1}{40} + i \frac{79}{40}

x = \frac{- ( - 1 ) - 79 i}{2 \cdot 20} : \frac{1}{40} - i \frac{79}{40}

The solutions to the quadratic equation are:

x = \frac{1}{40} + i \frac{79}{40}, x = \frac{1}{40} - i \frac{79}{40}

\frac{4}{9} + \frac{6}{18} + \frac{5}{3}

7 z \cdot (2 x - y) + 2 x - y

4 x - 9 = 3.5 x - 9

5 \cdot 12

What are the solutions to the equation 20x^2+1=x ?
The solutions to the equation 20x^2+1=x are x= 1/40+i(sqrt(79))/(40),x= 1/40-i(sqrt(79))/(40)
Find the zeros of 20x^2+1=x
The zeros of 20x^2+1=x are x= 1/40+i(sqrt(79))/(40),x= 1/40-i(sqrt(79))/(40)