Step 1: Let the greater number be $x$ and the smaller number be $y$, where $x,y>0$.
Step 2: From the first condition (difference of squares is 180),
$$x^{2}-y^{2}=180 \quad \cdots (i)$$
Step 3: From the second condition (square of smaller $=8\times$ greater),
$$y^{2}=8x \quad \cdots (ii)$$
Step 4: Substitute (ii) into (i).
$$x^{2}-8x=180$$
Step 5: Write it as a standard quadratic equation.
$$x^{2}-8x-180=0$$
Step 6: Factorise by splitting the middle term ($-18+10=-8$ and $-18\times 10=-180$).
$$x^{2}-18x+10x-180=0$$
$$x(x-18)+10(x-18)=0 \;\Rightarrow\; (x-18)(x+10)=0$$
Step 7: So $x=18$ or $x=-10$. Since the numbers are positive, we reject $x=-10$.
$$x=18$$
Step 8: Substitute $x=18$ in (ii).
$$y^{2}=8(18)=144 \;\Rightarrow\; y=12 \quad (y>0)$$
Step 9: Verification: $18^{2}-12^{2}=324-144=180$ ✓ and $12^{2}=144=8\times 18$ ✓
Hence the two numbers are $18$ and $12$.