Find the range of values of $c$ for which $x(10-x)<c^2$

I've manipulated the equation to end up with $(x-5)^2 + (c+5)(c-5)<0$, but I'm unsure how to proceed from here. Any help will be appreciated thanks!

$$10x-x^2=25-(x-5)^2\le25$$

So we need $$c^2>25$$ $\iff c>5$ or $c<-5$

$$x(10-x)\leq\left(\frac{x+10-x}{2}\right)^2=25,$$ which says $c^2>25$, which gives $$\mathbb R\setminus[-5,5]$$

Your question is essentially asking for the maximum value of $x(10-x)$ over $\mathbb{R}$, and since the function $f(x)=x(10-x)$ is concave and symmetric with respect to $x=5$ (i.e. $f(5-z)=f(5+z)$) such maximum is attained at $x=5$. Equivalently, $x(10-x)$ is non-negative only over $[0,10]$ and for any $x$ in such interval the AM-GM inequality grants $x(10-x)\leq\left(\frac{x+10-x}{2}\right)^2=25.$

Another attempt , pedestrian :

$y=x(10-x) = - (x - 5)^2 +25$ is a

parabola opening downward, vertex at $(5,25)$, with maximum $y=25$ (why?).

Choose: $c^2 >25$, i.e. $|5| >5.$