Solve for C C(t)=t^2-2t-4

$C (t) = t^{2} - 2 t - 4$

Step 1

Divide each term in $C t = t^{2} - 2 t - 4$ by $t$ .

$\frac{C t}{t} = \frac{t^{2}}{t} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Step 2

Simplify the left side.

Cancel the common factor of $t$ .

Cancel the common factor.

$\frac{C t}{t} = \frac{t^{2}}{t} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Divide $C$ by $1$ .

$C = \frac{t^{2}}{t} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Step 3

Simplify the right side.

Simplify each term.

Cancel the common factor of $t^{2}$ and $t$ .

Factor $t$ out of $t^{2}$ .

$C = \frac{t \cdot t}{t} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Cancel the common factors.

Raise $t$ to the power of $1$ .

$C = \frac{t \cdot t}{t^{1}} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Factor $t$ out of $t^{1}$ .

$C = \frac{t \cdot t}{t \cdot 1} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Cancel the common factor.

$C = \frac{t \cdot t}{t \cdot 1} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Rewrite the expression.

$C = \frac{t}{1} + \frac{- 2 t}{t} + \frac{- 4}{t}$

Divide $t$ by $1$ .

$C = t + \frac{- 2 t}{t} + \frac{- 4}{t}$

Cancel the common factor of $t$ .

Cancel the common factor.

$C = t + \frac{- 2 t}{t} + \frac{- 4}{t}$

Divide $- 2$ by $1$ .

$C = t - 2 + \frac{- 4}{t}$

Move the negative in front of the fraction.

$C = t - 2 - \frac{4}{t}$