Solve -5*1t+1/2gt^2=105

$$- 5 \cdot 1 t + \frac { 1 } { 2 } g t ^ { 2 } = 105$$

Solve for g

$$g=\frac{10\left(t+21\right)}{t^{2}}$$

$$t\neq 0$$

Steps for Solving Linear Equation

Multiply -5 and 1 to get -5.

$$-5t+\frac{1}{2}gt^{2}=105$$

Add 5t to both sides.

$$\frac{1}{2}gt^{2}=105+5t$$

The equation is in standard form.

$$\frac{t^{2}}{2}g=5t+105$$

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

$$\frac{2\times \frac{t^{2}}{2}g}{t^{2}}=\frac{2\left(5t+105\right)}{t^{2}}$$

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

$$g=\frac{2\left(5t+105\right)}{t^{2}}$$

Divide 105+5t by \frac{1}{2}t^{2}.

$$g=\frac{10\left(t+21\right)}{t^{2}}$$

Solve for t (complex solution)

$$\left\{\begin{matrix}t=\frac{\sqrt{210g+25}+5}{g}\text{; }t=\frac{-\sqrt{210g+25}+5}{g}\text{, }&g\neq 0\\t=-21\text{, }&g=0\end{matrix}\right.$$

Solve for t

$$\left\{\begin{matrix}t=\frac{\sqrt{210g+25}+5}{g}\text{; }t=\frac{-\sqrt{210g+25}+5}{g}\text{, }&g\neq 0\text{ and }g\geq -\frac{5}{42}\\t=-21\text{, }&g=0\end{matrix}\right.$$