The given equation $(x-2{)}^2+1=2x-3$

$LHS=(x–2{)}^2+1=x^2-4x+4+1=x^2-4x+5$

Therefore, $(x–2{)}^2+1=2x–3$ can be rewritten as $x^2–4x+5=2x–3$

i.e., $x^2–6x+8=0$

It is of the form $a{x}^2+bx+c=0$.

Therefore, the given equation is a quadratic equation.