解:因為 $a=2,b=-3,c=-3$ 所以 $b2-4ac$ $=(-3)2-4×2×(-3)=33$ x= $\frac{-b±\sqrt{b2-4ac} }{2a}$ = $\frac{-(-3)±\sqrt{33}}{2×2}$ = $\frac{3±\sqrt{33}}{2×2}$ ${x}_{1}=\frac{3+\sqrt{33}}{4},{x}_{2}=\frac{3-\sqrt{33}}{4}$
解:因為 $a=4,b=-1,c=-1$ $所以 b2-4ac$ $=(-1)2-4×4×(-1)=17$ x= $\frac{-b±\sqrt{b2-4ac} }{2a}$ = $\frac{1±\sqrt{17}}{2×4}$ = $\frac{1±\sqrt{17}}{8}$ ${x}_{1}=\frac{1+\sqrt{17}}{8},{x}_{2}=\frac{1-\sqrt{17}}{8}$ 解: $a=2,b=5,c=1$ $b2-4ac$ $=52-4×2×1=17$ y= $\frac{-b±\sqrt{b2-4ac} }{2a}$ = $\frac{-5±\sqrt{17}}{2×2}$ = $\frac{-5±\sqrt{17}}{4}$ ${y}_{1}=\frac{-5+\sqrt{17}}{4},{y}_{2}=\frac{-5-\sqrt{17}}{4}$
解: $a=3,b=-9,c=5$ $所以 b2-4ac$ $=(-9)2-4×3×5=21$ x= $\frac{-b±\sqrt{b2-4ac} }{2a}$ = $\frac{9±\sqrt{21}}{2×3}$ = $\frac{9±\sqrt{21}}{6}$ ${x}_{1}=\frac{9+\sqrt{21}}{6},{x}_{2}=\frac{9-\sqrt{21}}{6}$
解: $(x-1)(x-2)=0$ $x-1=0或x-2=0$ ${x}_{1}=1,{x}_{2}=2$
解: $(2x-1)(x+1)=0$ $2x-1=0或x+1=0$ ${x}_{1}=\frac1 2,{x}_{2}=-1$
解: $(x-3)(x+1)=0$ $x-3=0或x+1=0$ ${x}_{1}=3,{x}_{2}=-1$
解: $(y-1)(y+4)=0$ $y-1=0或y+4=0$ ${y}_{1}=1,{y}_{2}=-4$
解: $(x+4)(x+5)=0$ $x+5=0或x+4=0$ ${x}_{1}=-5,{x}_{2}=-4$
解:(x+5)2=0 ${x}_{1}={x}_{2}=-5$
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