Given a functional equality prove that the function is never $0$

For a function $u:\mathbb{R}\to\mathbb{R}$ with $u(t)^2+u(t)=t^2+t+2$ , $u(0)=1$, Can we prove that the graph of $u$ never touches the $t$ axis ?

We can easly see that by adding $0.25$ to both sides of the given equality we get

$$\left(u(t)+0.5\right)^2=(t+0.5)^2+2\implies |u(t)+0.5|=\sqrt{(t+0.5)^2+2}$$

Furthermore:

$$\text{RHS}\neq0\iffu(t)+0.5\neq0$$

So $u(t)+0.5>0\lor u(t)+0.5<0 ,\forall t\in\mathbb{R}$ and since we have the value of the function at the point $0$ we get $u(t)>-0.5$

How can I continue? Is this the wrong way to approach this? Any help would be appreciated!