If $a{x}^2+bx+10=0$  does not have two distinct real roots, then the least value of $5a+b$. is

lt is given that $a{x}^2+bx+10=0$does not have two distinct real roots.

$\therefore b^2-40a\le 0\Rightarrow a\ge \frac{b^2}{40}$

Let $y=5a+b$ Then,

$\begin{array}{l}y=5\times \frac{b^2}{40}+b=\frac{b^2+8b}{8}=\frac{1}{8}\left(b^2+8b\right)\\ y=\frac{1}{8}(b+4{)}^2-2\ge -2\end{array}$

Hence, the least value of 5a+b  is  -2.