#### Answer

$2$ solutions if $-6.25\lt c$, one solution if $-6.25=c$, and no solution if $-6.25\gt c$.

#### Work Step by Step

The discriminant is $D=b^2-4ac$. If $D\gt0$ we have $2$, if $D=0$ we have one, and if $D\lt0$ we have no solution.
Hence here $D=100+16c$. $100+16c\gt0\\100\gt-16c\\-6.25\lt c.$
$100+16c=0\\100=-16c\\-6.25= c.$
$100+16c\lt0\\100\lt-16c\\-6.25\gt c.$
Thus we have $2$ solutions if $-6.25\lt c$, one solution if $-6.25=c$, and no solution if $-6.25\gt c$.