Solve C(t)=t^2-2t+5quadyS(t)=3t-4

$$C ( t ) = t ^ { 2 } - 2 t + 5 \quad y S ( t ) = 3 t - 4$$

Solve for S (complex solution)

$$\left\{\begin{matrix}S=-\frac{\left(t-4\right)\left(t-1\right)}{5ty}\text{, }&t\neq 0\text{ and }y\neq 0\\S\in \mathrm{C}\text{, }&\left(t=4\text{ or }t=1\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract t^{2} from both sides.

$$-2t+5ySt=3t-4-t^{2}$$

Add 2t to both sides.

$$5ySt=3t-4-t^{2}+2t$$

Combine 3t and 2t to get 5t.

$$5ySt=5t-4-t^{2}$$

The equation is in standard form.

$$5tyS=-t^{2}+5t-4$$

Divide both sides by 5yt.

$$\frac{5tyS}{5ty}=\frac{\left(1-t\right)\left(t-4\right)}{5ty}$$

Dividing by 5yt undoes the multiplication by 5yt.

$$S=\frac{\left(1-t\right)\left(t-4\right)}{5ty}$$

Solve for S

$$\left\{\begin{matrix}S=-\frac{\left(t-4\right)\left(t-1\right)}{5ty}\text{, }&t\neq 0\text{ and }y\neq 0\\S\in \mathrm{R}\text{, }&\left(t=4\text{ or }t=1\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract t^{2} from both sides.

$$-2t+5ySt=3t-4-t^{2}$$

Add 2t to both sides.

$$5ySt=3t-4-t^{2}+2t$$

Combine 3t and 2t to get 5t.

$$5ySt=5t-4-t^{2}$$

The equation is in standard form.

$$5tyS=-t^{2}+5t-4$$

Divide both sides by 5yt.

$$\frac{5tyS}{5ty}=\frac{\left(1-t\right)\left(t-4\right)}{5ty}$$

Dividing by 5yt undoes the multiplication by 5yt.

$$S=\frac{\left(1-t\right)\left(t-4\right)}{5ty}$$

Solve for t (complex solution)

$$t=\frac{\sqrt{9+25\left(Sy\right)^{2}-50Sy}-5Sy+5}{2}$$

$$t=\frac{-\sqrt{9+25\left(Sy\right)^{2}-50Sy}-5Sy+5}{2}$$

Solve for t

$$t=\frac{\sqrt{9+25\left(Sy\right)^{2}-50Sy}-5Sy+5}{2}$$

$$t=\frac{-\sqrt{9+25\left(Sy\right)^{2}-50Sy}-5Sy+5}{2}\text{, }S=0\text{ or }y\geq \frac{4|S|}{5S^{2}}+\frac{1}{S}\text{ or }y\leq -\frac{4|S|}{5S^{2}}+\frac{1}{S}$$