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Solution 2.3:2c

From Förberedande kurs i matematik 1

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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.

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