What are the solutions to the equation 0=0.5F(1-F/(3000))-100 ?

The solutions to the equation 0=0.5F(1-F/(3000))-100 are F=300(-sqrt(55/3)+5),F=300(sqrt(55/3)+5)

Find the zeros of 0=0.5F(1-F/(3000))-100

The zeros of 0=0.5F(1-F/(3000))-100 are F=300(-sqrt(55/3)+5),F=300(sqrt(55/3)+5)

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0 = 0.5 F (1 - \frac{F}{3000}) - 100

F = 300 (- \frac{55}{3} + 5), F = 300 (\frac{55}{3} + 5)

Decimal Notation

F = 215.47674 \dots, F = 2784.52325 \dots

Solution steps

0 = 0.5 F (1 - \frac{F}{3000}) - 100

Multiply both sides by

10

0 = 5 F (1 - \frac{F}{3000}) - 1000

Expand

5 F (1 - \frac{F}{3000}) - 1000 : 5 F - \frac{F ^{2}}{600} - 1000

0 = 5 F - \frac{F ^{2}}{600} - 1000

Switch sides

5 F - \frac{F ^{2}}{600} - 1000 = 0

Write in the standard form

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

- \frac{F ^{2}}{600} + 5 F - 1000 = 0

Solve with the quadratic formula

F_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - \frac{1}{600} ) ( - 1000 )}{2 ( - \frac{1}{600} )}

5^{2} - 4 (- \frac{1}{600}) (- 1000) = \frac{55}{3}

F_{1, 2} = \frac{- 5 \pm \frac{55}{3}}{2 ( - \frac{1}{600} )}

Separate the solutions

F_{1} = \frac{- 5 + \frac{55}{3}}{2 ( - \frac{1}{600} )}, F_{2} = \frac{- 5 - \frac{55}{3}}{2 ( - \frac{1}{600} )}

F = \frac{- 5 + \frac{55}{3}}{2 ( - \frac{1}{600} )} : 300 (- \frac{55}{3} + 5)

F = \frac{- 5 - \frac{55}{3}}{2 ( - \frac{1}{600} )} : 300 (\frac{55}{3} + 5)

The solutions to the quadratic equation are:

F = 300 (- \frac{55}{3} + 5), F = 300 (\frac{55}{3} + 5)

- 3 t^{2} = - 8 t + 6

0.25 x^{3} - 0.25 x (x^{2} - 6 x - 30) = 21

y, x^{2} - 5 x y + 3 y^{2} = 7

6 x^{2} + 19 x + 15 = 0

- 12 q^{2} + 3 q + 7 = 0

What are the solutions to the equation 0=0.5F(1-F/(3000))-100 ?
The solutions to the equation 0=0.5F(1-F/(3000))-100 are F=300(-sqrt(55/3)+5),F=300(sqrt(55/3)+5)
Find the zeros of 0=0.5F(1-F/(3000))-100
The zeros of 0=0.5F(1-F/(3000))-100 are F=300(-sqrt(55/3)+5),F=300(sqrt(55/3)+5)