[tex]ax^2+bx+c=0[/tex]
if m and n are zeros, then:
[tex]m+n=\dfrac{-b}{a}\ and\ mn=\dfrac{c}{a}[/tex]
We have:
[tex]ax^2-5x+c=0\to b=-5\\\\m+n=mn=10[/tex]
substitute
[tex]m+n=\dfrac{-(-5)}{a};\ mn=\dfrac{c}{a}\\\\ \left\{\begin{array}{ccc}\dfrac{5}{a}=10&\to a=\dfrac{1}{2}\\\\\dfrac{c}{a}=10\end{array}\right[/tex]
substitute the value of a to the second equation:
[tex]\dfrac{c}{\frac{1}{2}}=10\\\\2c=10\ \ \ |:2\\\\c=5[/tex]
Answer: [tex]\boxed{a=\dfrac{1}{2};\ c=5}[/tex]
