Solving simple circuit problem
You are told that the current with R is more than the current with R and some added resistance by a specified amount. So write this very simple English statement down as an equation:

$$\begin{align*}
I_{R}-I_{R+4\:\Omega}&amp;=500\:\textrm{mA}\\\\
\end{align*}$$

But you also know Ohm's law. So, just replace the currents by the obvious fractions to find:

$$\begin{align*}
I_{R}-I_{R+4\:\Omega}&amp;=500\:\textrm{mA}\\\\
\frac{12\:\textrm{V}}{R}-\frac{12\:\textrm{V}}{R+4\:\Omega}&amp;=500\:\textrm{mA}
\end{align*}$$

And solve for \$R\$.

Hopefully, you can see how you start by just writing down what you read with the simplest possible quantitative statement about it, and then follow it up by filling in more details until you can see something that looks as though it is solvable. Then you solve it.



$$\begin{align*}
\frac{12\:\textrm{V}}{R}-\frac{12\:\textrm{V}}{R+4\:\Omega}&amp;=500\:\textrm{mA} \\\\
12\:\textrm{V}\cdot\left(\frac{1}{R}-\frac{1}{R+4\:\Omega}\right)&amp;=500\:\textrm{mA} \\\\
12\:\textrm{V}\cdot \left(\frac{1}{R}-\frac{1}{R+4\:\Omega}\right)\cdot R\cdot \left(R+4\:\Omega\right)&amp;=500\:\textrm{mA}\cdot R\cdot \left(R+4\:\Omega\right) \\\\
12\:\textrm{V}\cdot \left(\frac{R\cdot \left(R+4\:\Omega\right)}{R}-\frac{R\cdot \left(R+4\:\Omega\right)}{R+4\:\Omega}\right)&amp;=500\:\textrm{mA}\cdot R\cdot \left(R+4\:\Omega\right) \\\\
12\:\textrm{V}\cdot \left(R+4\:\Omega-R\right)&amp;=500\:\textrm{mA}\cdot R\cdot \left(R+4\:\Omega\right) \\\\
500\:\textrm{mA}\cdot R\cdot \left(R+4\:\Omega\right)&amp;=12\:\textrm{V}\cdot 4\:\Omega \\\\
R\cdot \left(R+4\:\Omega\right)&amp;=\frac{12\:\textrm{V}\cdot 4\:\Omega}{500\:\textrm{mA}}= 96\:\Omega^2 \\\\
R^2+4\:\Omega\cdot R&amp;= 96\:\Omega^2 \\\\
R^2+4\:\Omega\cdot R&amp;- 96\:\Omega^2 = 0
\end{align*}$$

That's a quadratic equation form (\$a\:R^2+b\:R+c=0\$), where \$a=1\$, \$b=4\$, and \$c=-96\$. That is solvable:

$$\begin{align*}
R&amp;=\frac{-b\pm\sqrt{b^2-4\: a\: c}}{2\: a}\\\\
&amp;=\frac{-4\pm\sqrt{4^2-4\cdot 1\cdot -96}}{2\cdot 1}\\\\
&amp;=\frac{-4\pm\sqrt{400}}{2}\\\\
&amp;=\frac{-4\pm 20}{2}\\\\
&amp;=-2\pm 10\\\\
\end{align*}$$

And there is only one realistic value there, which is \$R=8\:\Omega\$.