What are the solutions to the equation x^2+1.1x-0.3=0 ?

The solutions to the equation x^2+1.1x-0.3=0 are x=(-11+sqrt(241))/(20),x=(-11-sqrt(241))/(20)

Find the zeros of x^2+1.1x-0.3=0

The zeros of x^2+1.1x-0.3=0 are x=(-11+sqrt(241))/(20),x=(-11-sqrt(241))/(20)

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x^{2} + 1.1 x - 0.3 = 0

x = \frac{- 11 + 241}{20}, x = \frac{- 11 - 241}{20}

Decimal Notation

x = 0.22620 \dots, x = - 1.32620 \dots

Solution steps

x^{2} + 1.1 x - 0.3 = 0

Multiply both sides by

10

10 x^{2} + 11 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 10 ( - 3 )}{2 \cdot 10}

1 1^{2} - 4 \cdot 10 (- 3) = 241

x_{1, 2} = \frac{- 11 \pm 241}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 11 + 241}{2 \cdot 10}, x_{2} = \frac{- 11 - 241}{2 \cdot 10}

x = \frac{- 11 + 241}{2 \cdot 10} : \frac{- 11 + 241}{20}

x = \frac{- 11 - 241}{2 \cdot 10} : \frac{- 11 - 241}{20}

The solutions to the quadratic equation are:

x = \frac{- 11 + 241}{20}, x = \frac{- 11 - 241}{20}

2 π 6

x^{2} - 3 x > 3 x - 9

2 x^{2} + 14 x = 0

16 x \leq x^{3}

u + 1.8 = 8.26

What are the solutions to the equation x^2+1.1x-0.3=0 ?
The solutions to the equation x^2+1.1x-0.3=0 are x=(-11+sqrt(241))/(20),x=(-11-sqrt(241))/(20)
Find the zeros of x^2+1.1x-0.3=0
The zeros of x^2+1.1x-0.3=0 are x=(-11+sqrt(241))/(20),x=(-11-sqrt(241))/(20)