If $(2x-k{)}^2=4x^2-20x+25$, what is the value of $k$?

$(2x-k{)}^2=(2x{)}^2-2(2x)(k)+k^2=4x^2-4kx+k^2$. Matching the middle term: $4k=20$, so $k=5$ — and $k^2=25$ matches the constant, confirming it. Reading the constant 25 straight off as $k$ skips the square root; solving $2k=20$ instead of $4k=20$ gives 10.