What are the solutions to the equation C(t)=-0.0002t^2+0.06t,0<= t<= 240 ?

The solutions to the equation C(t)=-0.0002t^2+0.06t,0<= t<= 240 are t=0,t=(-C+0.06)/(0.0002)

Find the zeros of C(t)=-0.0002t^2+0.06t,0<= t<= 240

The zeros of C(t)=-0.0002t^2+0.06t,0<= t<= 240 are t=0,t=(-C+0.06)/(0.0002)

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The solutions to the equation C(t)=-0.0002t^2+0.06t,0<= t<= 240 are t=0,t=(-C+0.06)/(0.0002)

The zeros of C(t)=-0.0002t^2+0.06t,0<= t<= 240 are t=0,t=(-C+0.06)/(0.0002)

C (t) = - 0.0002 t^{2} + 0.06 t, 0 \leq t \leq 240

t = 0, t = \frac{- C + 0.06}{0.0002}

Solution steps

C (t) = - 0.0002 t^{2} + 0.06 t, 0 \leq t \leq 240

Refine

Ct = - 0.0002 t^{2} + 0.06 t

Switch sides

- 0.0002 t^{2} + 0.06 t = Ct

Move

Ct

to the left side

- 0.0002 t^{2} + 0.06 t - Ct = 0

Factor

- 0.0002 t^{2} + 0.06 t - Ct : - t (0.0002 t + C - 0.06)

- t (0.0002 t + C - 0.06) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

t = 0 or 0.0002 t + C - 0.06 = 0

Solve

0.0002 t + C - 0.06 = 0 : t = \frac{- C + 0.06}{0.0002}

The solutions to the quadratic equation are:

t = 0, t = \frac{- C + 0.06}{0.0002}

Check constrains

t = 0, t = \frac{- C + 0.06}{0.0002}

110.25 - 17.5 y + 7 y^{2} = 120

x, x^{2} + 4 y^{2} + z^{2} = 9

5 - 11 x^{2} = - 8 x

3 x^{2} - 8 x = 8

4.69742 x^{2} + 53.20224 x - 4738.618 = 0

What are the solutions to the equation C(t)=-0.0002t^2+0.06t,0<= t<= 240 ?
The solutions to the equation C(t)=-0.0002t^2+0.06t,0<= t<= 240 are t=0,t=(-C+0.06)/(0.0002)
Find the zeros of C(t)=-0.0002t^2+0.06t,0<= t<= 240
The zeros of C(t)=-0.0002t^2+0.06t,0<= t<= 240 are t=0,t=(-C+0.06)/(0.0002)