What are the solutions to the equation s^2+(2+0.1)s+0.1=0 ?

The solutions to the equation s^2+(2+0.1)s+0.1=0 are s=(-21+sqrt(401))/(20),s=(-21-sqrt(401))/(20)

Find the zeros of s^2+(2+0.1)s+0.1=0

The zeros of s^2+(2+0.1)s+0.1=0 are s=(-21+sqrt(401))/(20),s=(-21-sqrt(401))/(20)

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s^{2} + (2 + 0.1) s + 0.1 = 0

s = \frac{- 21 + 401}{20}, s = \frac{- 21 - 401}{20}

Decimal Notation

s = - 0.04875 \dots, s = - 2.05124 \dots

Solution steps

s^{2} + (2 + 0.1) s + 0.1 = 0

Multiply both sides by

10

10 s^{2} + 21 s + 1 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 21 \pm 2 1 ^{2} - 4 \cdot 10 \cdot 1}{2 \cdot 10}

2 1^{2} - 4 \cdot 10 \cdot 1 = 401

s_{1, 2} = \frac{- 21 \pm 401}{2 \cdot 10}

Separate the solutions

s_{1} = \frac{- 21 + 401}{2 \cdot 10}, s_{2} = \frac{- 21 - 401}{2 \cdot 10}

s = \frac{- 21 + 401}{2 \cdot 10} : \frac{- 21 + 401}{20}

s = \frac{- 21 - 401}{2 \cdot 10} : \frac{- 21 - 401}{20}

The solutions to the quadratic equation are:

s = \frac{- 21 + 401}{20}, s = \frac{- 21 - 401}{20}

y^{2} - 20 = 101

(23)^{2} + 2^{2} = x^{2}

\frac{x ^{2}}{25} - \frac{x ^{2}}{144} = 1

- 16 x^{2} + 36 x - 14 = 0

(2 x + 1)^{2} = 4

What are the solutions to the equation s^2+(2+0.1)s+0.1=0 ?
The solutions to the equation s^2+(2+0.1)s+0.1=0 are s=(-21+sqrt(401))/(20),s=(-21-sqrt(401))/(20)
Find the zeros of s^2+(2+0.1)s+0.1=0
The zeros of s^2+(2+0.1)s+0.1=0 are s=(-21+sqrt(401))/(20),s=(-21-sqrt(401))/(20)