What are the solutions to the equation s^2-0.1s+1=0 ?

The solutions to the equation s^2-0.1s+1=0 are s= 1/20+i(sqrt(399))/(20),s= 1/20-i(sqrt(399))/(20)

Find the zeros of s^2-0.1s+1=0

The zeros of s^2-0.1s+1=0 are s= 1/20+i(sqrt(399))/(20),s= 1/20-i(sqrt(399))/(20)

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s^{2} - 0.1 s + 1 = 0

s = \frac{1}{20} + i \frac{399}{20}, s = \frac{1}{20} - i \frac{399}{20}

Solution steps

s^{2} - 0.1 s + 1 = 0

Multiply both sides by

10

10 s^{2} - s + 10 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 10 \cdot 10}{2 \cdot 10}

Simplify

(- 1)^{2} - 4 \cdot 10 \cdot 10 : 399 i

s_{1, 2} = \frac{- ( - 1 ) \pm 399 i}{2 \cdot 10}

Separate the solutions

s_{1} = \frac{- ( - 1 ) + 399 i}{2 \cdot 10}, s_{2} = \frac{- ( - 1 ) - 399 i}{2 \cdot 10}

s = \frac{- ( - 1 ) + 399 i}{2 \cdot 10} : \frac{1}{20} + i \frac{399}{20}

s = \frac{- ( - 1 ) - 399 i}{2 \cdot 10} : \frac{1}{20} - i \frac{399}{20}

The solutions to the quadratic equation are:

s = \frac{1}{20} + i \frac{399}{20}, s = \frac{1}{20} - i \frac{399}{20}

x^{2} + 8 x + 16 = 1

x, f = \frac{5}{3} m^{4} x^{2} - 7 m^{3} x - 3 m

u^{2} - 7 u - 18 = 0

x^{2} + 3 x = x + 7

x^{2} = 1 3^{2} - 1 2^{2}

What are the solutions to the equation s^2-0.1s+1=0 ?
The solutions to the equation s^2-0.1s+1=0 are s= 1/20+i(sqrt(399))/(20),s= 1/20-i(sqrt(399))/(20)
Find the zeros of s^2-0.1s+1=0
The zeros of s^2-0.1s+1=0 are s= 1/20+i(sqrt(399))/(20),s= 1/20-i(sqrt(399))/(20)