Solve vf^2=vi^2+20Deltax

$$v f ^ { 2 } = v i ^ { 2 } + 20 \Delta x$$

Solve for v

$$\left\{\begin{matrix}v=\frac{20x\Delta }{f^{2}+1}\text{, }&f\neq -i\text{ and }f\neq i\\v\in \mathrm{C}\text{, }&\left(x=0\text{ and }f=i\right)\text{ or }\left(x=0\text{ and }f=-i\right)\text{ or }\left(\Delta =0\text{ and }f=i\right)\text{ or }\left(\Delta =0\text{ and }f=-i\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate i to the power of 2 and get -1.

$$vf^{2}=v\left(-1\right)+20\Delta x$$

Subtract v\left(-1\right) from both sides.

$$vf^{2}-v\left(-1\right)=20\Delta x$$

Multiply -1 and -1 to get 1.

$$vf^{2}+v=20\Delta x$$

Combine all terms containing v.

$$\left(f^{2}+1\right)v=20\Delta x$$

The equation is in standard form.

$$\left(f^{2}+1\right)v=20x\Delta$$

Divide both sides by f^{2}+1.

$$\frac{\left(f^{2}+1\right)v}{f^{2}+1}=\frac{20x\Delta }{f^{2}+1}$$

Dividing by f^{2}+1 undoes the multiplication by f^{2}+1.

$$v=\frac{20x\Delta }{f^{2}+1}$$

Solve for f

$$\left\{\begin{matrix}f=-\sqrt{\frac{20x\Delta }{v}-1}\text{; }f=\sqrt{\frac{20x\Delta }{v}-1}\text{, }&v\neq 0\\f\in \mathrm{C}\text{, }&\left(x=0\text{ or }\Delta =0\right)\text{ and }v=0\end{matrix}\right.$$