Mathematics - Quadratic Equation Question with Solution | TestHub

The quadratic expression $x^2+2ax+10-3a>0$ for all real $x$ implies its discriminant must be negative.

For $A{x}^2+Bx+C>0$, we need $A>0$ and $D<0$. Here $A=1>0$.

Discriminant $D=(2a{)}^2-4(1)(10-3a)<0$.

$4a^2-40+12a<0$.

$a^2+3a-10<0$.

Factoring, $(a+5)(a-2)<0$.

This inequality holds when $-5<a<2$.