The first-order ODE $$\frac{\dd y}{\dd t}=\frac{3 t^2+4 t -1}{2y}$$ can be rewritten in differential form as $$2y\,\dd y=(3 t^2+4 t -1)\,\dd t$$ The left- and right-hand sides can be replaced by $\dd(y^2)$ and $\dd(t^3+2 t^2-t)$, respectively. After all, $y^2$ and $t^3+2 t^2-t$ are antiderivatives of $y$ and $3 t^2+4 t -1$, respectively. So we have an equality between two differentials: $$\dd(y^2)=\dd(t^3+2 t^2-t)$$ The functions ‘behind the d's’ are thus equal to each other up to a constant: $$y^2=t^3+2 t^2-t+c$$ for some constant $c$. The explicit solution is: $$y=\sqrt{t^3+2 t^2-t+c}\quad \text{or} \quad y=-\sqrt{t^3+2 t^2-t+c}$$

For those who find the final steps in the processing of differentials too formal or are not familar with differentials: you can also add to the left- and right-hand side of the $$2y\,\dd y=(3 t^2+4 t -1)\,\dd t$$ the integral symbol: $$\int 2y\,\dd y=\int (3 t^2+4 t -1)\,\dd t$$ Now you just continue with computing these integrals. So: $$y^2=t^3+2 t^2-t+c$$ where we have combined the integration constants into a single constant $c$ and have placed it on the right-hand side of the equation.