Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2-m\right)^{2}$.
$$4-4m+m^{2}-4\times 2\times 2<0$$
Multiply $4$ and $2$ to get $8$.
$$4-4m+m^{2}-8\times 2<0$$
Multiply $8$ and $2$ to get $16$.
$$4-4m+m^{2}-16<0$$
Subtract $16$ from $4$ to get $-12$.
$$-12-4m+m^{2}<0$$
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$-12-4m+m^{2}=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-4$ for $b$, and $-12$ for $c$ in the quadratic formula.
