From Förberedande kurs i matematik 1

We start by completing the square of the left-hand side:

\displaystyle \begin{align} & y^{2}+3y+4=\left( y+\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \\ & =\left( y+\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4}=\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4} \\ \end{align}

The equation is then

\displaystyle \left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0

The first term \displaystyle \left( y+\frac{3}{2} \right)^{2} is always greater than or equal to zero because it is a square and \displaystyle \frac{7}{4} is a positive number. This means that the left hand side cannot be zero, regardless of how \displaystyle y is chosen. The equation has no solution.