We start by completing the square of the left-hand side:
y2+3y+4=y+232−232+4=y+232−49+416=y+232+47
The equation is then
y+232+47=0
The first term
y+232
is always greater than or equal to zero because it is a square and
47
is a positive number. This means that the left hand side cannot be zero, regardless of how
y
is chosen. The equation has no solution.
y+23
2−