What are the solutions to the equation 117x^2+1=18x ?

The solutions to the equation 117x^2+1=18x are x= 1/13+i 2/39 ,x= 1/13-i 2/39

Find the zeros of 117x^2+1=18x

The zeros of 117x^2+1=18x are x= 1/13+i 2/39 ,x= 1/13-i 2/39

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117 x^{2} + 1 = 18 x

x = \frac{1}{13} + i \frac{2}{39}, x = \frac{1}{13} - i \frac{2}{39}

Solution steps

117 x^{2} + 1 = 18 x

Move

18 x

to the left side

117 x^{2} + 1 - 18 x = 0

Write in the standard form

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

117 x^{2} - 18 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 117 \cdot 1}{2 \cdot 117}

Simplify

(- 18)^{2} - 4 \cdot 117 \cdot 1 : 12 i

x_{1, 2} = \frac{- ( - 18 ) \pm 12 i}{2 \cdot 117}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 12 i}{2 \cdot 117}, x_{2} = \frac{- ( - 18 ) - 12 i}{2 \cdot 117}

x = \frac{- ( - 18 ) + 12 i}{2 \cdot 117} : \frac{1}{13} + i \frac{2}{39}

x = \frac{- ( - 18 ) - 12 i}{2 \cdot 117} : \frac{1}{13} - i \frac{2}{39}

The solutions to the quadratic equation are:

x = \frac{1}{13} + i \frac{2}{39}, x = \frac{1}{13} - i \frac{2}{39}

37.5 X^{2} = 16, 4 + \frac{0.72}{X}

y, 0 = 0.05 y - 0.0005 y^{2}

(x + 3) (x + 4) = 7 (x + 2)

x^{2} - 300 x - 22500 = 0

6 x^{2} - 8 x - 7 = 0

What are the solutions to the equation 117x^2+1=18x ?
The solutions to the equation 117x^2+1=18x are x= 1/13+i 2/39 ,x= 1/13-i 2/39
Find the zeros of 117x^2+1=18x
The zeros of 117x^2+1=18x are x= 1/13+i 2/39 ,x= 1/13-i 2/39