jmc

Since the root $5-4i$ is nonreal but the coefficients of the quadratic are real, the roots must form a conjugate pair. Therefore, the other root is $\overline{5-4i}=5+4i.$

To find the quadratic, we can note that the sum of the roots is $5-4i+5+4i=10$ and the product is $(5-4i)(5+4i)=25+16=41.$ Then by Vieta's formulas, we know that the quadratic $\boxed{x^2-10x+41}$ has $5-4i$ as a root.