What are the solutions to the equation b^2+110b-19.7*10^4=0 ?

The solutions to the equation b^2+110b-19.7*10^4=0 are b=5(3sqrt(889)-11),b=-5(11+3sqrt(889))

Find the zeros of b^2+110b-19.7*10^4=0

The zeros of b^2+110b-19.7*10^4=0 are b=5(3sqrt(889)-11),b=-5(11+3sqrt(889))

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b^{2} + 110 b - 19.7 \cdot 1 0^{4} = 0

b = 5 (3889 - 11), b = - 5 (11 + 3889)

Decimal Notation

b = 392.24154 \dots, b = - 502.24154 \dots

Solution steps

b^{2} + 110 b - 19.7 \cdot 1 0^{4} = 0

Multiply both sides by

10

10 b^{2} + 1100 b - 1970000 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- 1100 \pm 110 0 ^{2} - 4 \cdot 10 ( - 1970000 )}{2 \cdot 10}

110 0^{2} - 4 \cdot 10 (- 1970000) = 300889

b_{1, 2} = \frac{- 1100 \pm 300 889}{2 \cdot 10}

Separate the solutions

b_{1} = \frac{- 1100 + 300 889}{2 \cdot 10}, b_{2} = \frac{- 1100 - 300 889}{2 \cdot 10}

b = \frac{- 1100 + 300 889}{2 \cdot 10} : 5 (3889 - 11)

b = \frac{- 1100 - 300 889}{2 \cdot 10} : - 5 (11 + 3889)

The solutions to the quadratic equation are:

b = 5 (3889 - 11), b = - 5 (11 + 3889)

12 0^{2} + 5 0^{2} = x^{2}

4 y^{2} + 6 y + 2 = 0

- 2 x^{2} + 11 x + 12 = 0

x^{2} - 11 x = x - 36

x (x + 8) = 12

What are the solutions to the equation b^2+110b-19.7*10^4=0 ?
The solutions to the equation b^2+110b-19.7*10^4=0 are b=5(3sqrt(889)-11),b=-5(11+3sqrt(889))
Find the zeros of b^2+110b-19.7*10^4=0
The zeros of b^2+110b-19.7*10^4=0 are b=5(3sqrt(889)-11),b=-5(11+3sqrt(889))