VicKmath7 wrote: Find all real numbers $q$, such that for all real $p \geq 0$, the equation $x^2-2px+q^2+q-2=0$ has at least one real root in $(-1;0)$. Let $f(x)=x^2-2px+q^2+q-2$ Since $p\ge 0$ and one root in $(-1,0)$, the second root is $>0$ and so problem is equivalent to $f(-1)>0$ and $f(0)<0$ And so $q^2+q-2<0$ and $1+2p+q^2+q-2>0$ And so $0>q^2+q-2>-1-2p$ $\forall p\ge 0$ And so $0>q^2+q-2>-1$ And so $q\in(-2,1)$ and $q\notin\left[\frac{-1-\sqrt 5}2,\frac{-1+\sqrt 5}2\right]$ And so $\boxed{q\in\left(-2,\frac{-1-\sqrt 5}2\right)\cup\left(\frac{-1+\sqrt 5}2,1\right)}$