Solve S=ut-1/2Gt^2

$$S = u t - \frac { 1 } { 2 } G t ^ { 2 }$$

Solve for G (complex solution)

$$\left\{\begin{matrix}G=\frac{2\left(tu-S\right)}{t^{2}}\text{, }&t\neq 0\\G\in \mathrm{C}\text{, }&S=0\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$ut-\frac{1}{2}Gt^{2}=S$$

Subtract ut from both sides.

$$-\frac{1}{2}Gt^{2}=S-ut$$

The equation is in standard form.

$$\left(-\frac{t^{2}}{2}\right)G=S-tu$$

Divide both sides by -\frac{1}{2}t^{2}.

$$\frac{\left(-\frac{t^{2}}{2}\right)G}{-\frac{t^{2}}{2}}=\frac{S-tu}{-\frac{t^{2}}{2}}$$

Dividing by -\frac{1}{2}t^{2} undoes the multiplication by -\frac{1}{2}t^{2}.

$$G=\frac{S-tu}{-\frac{t^{2}}{2}}$$

Divide S-ut by -\frac{1}{2}t^{2}.

$$G=-\frac{2\left(S-tu\right)}{t^{2}}$$

Solve for G

$$\left\{\begin{matrix}G=\frac{2\left(tu-S\right)}{t^{2}}\text{, }&t\neq 0\\G\in \mathrm{R}\text{, }&S=0\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$ut-\frac{1}{2}Gt^{2}=S$$

Subtract ut from both sides.

$$-\frac{1}{2}Gt^{2}=S-ut$$

The equation is in standard form.

$$\left(-\frac{t^{2}}{2}\right)G=S-tu$$

Divide both sides by -\frac{1}{2}t^{2}.

$$\frac{\left(-\frac{t^{2}}{2}\right)G}{-\frac{t^{2}}{2}}=\frac{S-tu}{-\frac{t^{2}}{2}}$$

Dividing by -\frac{1}{2}t^{2} undoes the multiplication by -\frac{1}{2}t^{2}.

$$G=\frac{S-tu}{-\frac{t^{2}}{2}}$$

Divide S-tu by -\frac{1}{2}t^{2}.

$$G=-\frac{2\left(S-tu\right)}{t^{2}}$$

Solve for S

$$S=-\frac{t\left(Gt-2u\right)}{2}$$