What are the solutions to the equation 1-1.9x-0.1x^2=0 ?

The solutions to the equation 1-1.9x-0.1x^2=0 are x=-(19+sqrt(401))/2 ,x=(sqrt(401)-19)/2

Find the zeros of 1-1.9x-0.1x^2=0

The zeros of 1-1.9x-0.1x^2=0 are x=-(19+sqrt(401))/2 ,x=(sqrt(401)-19)/2

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1 - 1.9 x - 0.1 x^{2} = 0

x = - \frac{19 + 401}{2}, x = \frac{401 - 19}{2}

Decimal Notation

x = - 19.51249 \dots, x = 0.51249 \dots

Solution steps

1 - 1.9 x - 0.1 x^{2} = 0

Multiply both sides by

10

10 - 19 x - x^{2} = 0

Write in the standard form

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

- x^{2} - 19 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 ( - 1 ) \cdot 10}{2 ( - 1 )}

(- 19)^{2} - 4 (- 1) \cdot 10 = 401

x_{1, 2} = \frac{- ( - 19 ) \pm 401}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 401}{2 ( - 1 )}, x_{2} = \frac{- ( - 19 ) - 401}{2 ( - 1 )}

x = \frac{- ( - 19 ) + 401}{2 ( - 1 )} : - \frac{19 + 401}{2}

x = \frac{- ( - 19 ) - 401}{2 ( - 1 )} : \frac{401 - 19}{2}

The solutions to the quadratic equation are:

x = - \frac{19 + 401}{2}, x = \frac{401 - 19}{2}

u, E = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

(- 2 y)^{2} + y^{2} = 5

5 x^{2} + 25 = 0

9 x^{2} + 24 x - 7 = 0

2 x^{2} = 3 x - 10

What are the solutions to the equation 1-1.9x-0.1x^2=0 ?
The solutions to the equation 1-1.9x-0.1x^2=0 are x=-(19+sqrt(401))/2 ,x=(sqrt(401)-19)/2
Find the zeros of 1-1.9x-0.1x^2=0
The zeros of 1-1.9x-0.1x^2=0 are x=-(19+sqrt(401))/2 ,x=(sqrt(401)-19)/2