<s>[INST] How do I find the $x$ intercepts for $-x^2-3x+3$ How do I find the $x$ intercepts? $-x^2-3x+3$? 
I converted the function into vertex form but I am stuck at $3/4= -(x^2+1.5)$.
Can someone give me and idea what I can do? 
 [/INST] Alternatively, you can complete the square.
$x^2 + 3x - 3 = (x + \frac{3}{2})^2 - 3 - \frac{9}{4} = 0$. So, $(x + \frac{3}{2})^2 = \frac{21}{4} \Rightarrow x = -\frac{3}{2} \pm \frac {\sqrt{21}}2.$
Thus the x-intercepts are $(\frac{-3+\sqrt{21}}{2},0)$  and $(\frac{-3-\sqrt{21}}{2},0)$.
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