Using the quadratic formula why does y² + y + 1 = 0 equal a negative answer ??
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If you calculate the discriminant $\displaystyle D=1^2-4\cdot 1\cdot 1=1-4=-3$, this means there're no real zero's (only complex zero's), therefore there're no solutions in $\displaystyle \mathbb{R}$. Does this answer your question?
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