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解：?$x^2-4x = 2$?
?$a=1，$??$b=-4，$??$c=-2$?
?$b^2-4ac=(-4)^2-4×1×(-2)$?
?$=24＞0$?
?$x=\frac {4±\sqrt {24}}{2×1}$?
?$x_{1}=2+\sqrt {6}，$??$x_{2}=2-\sqrt {6}$?
解：?$\frac 52x^2+2x=1$?
?$5x^2+4x - 2 = 0$?
?$a = 5，$??$b = 4，$??$c=-2$?
?$b^2-4ac=4^2-4×5×(-2)$?
?$ = 56＞0$?
?$x=\frac {-4\pm \sqrt {56}}{10}$?
?$x_{1}=\frac {-2+\sqrt {14}}{5}，$??$x_{2}=\frac {-2-\sqrt {14}}{5}$?
解：?$p(p - 8)=16$?
?$p^2-8p - 16 = 0$?
?$a=1，$??$b=-8，$??$c=-16$?
?$b^2-4ac=(-8)^2-4×1×(-16)$?
?$=128＞0$?
?$p=\frac {8±\sqrt {128}}{2×1}=\frac {8±8\sqrt 2}2$?
?$p_{1}=4 + 4\sqrt {2}，$??$p_{2}=4 - 4\sqrt {2}$?
解：?$8y - 1 = 4y^2$?
?$4y^2-8y + 1 = 0$?
?$a = 4，$??$b=-8，$??$c = 1$?
?$b^2-4ac=(-8)^2-4×4×1$?
?$=48＞0$?
?$y=\frac {8\pm \sqrt {48}}{8}=\frac {2\pm \sqrt {3}}{2}$?
?$y_{1}=\frac {2+\sqrt {3}}{2}，$??$y_{2}=\frac {2-\sqrt {3}}{2}$?
解：在方程?$x^2-3x - 5 = 0$?中
?$a = 1，$??$b=-3，$??$c = -5，$?
?$ $?則?$?=(-3)^2-4×1×(-5)=29>0，$?
所以方程有兩個(gè)不相等的實(shí)數(shù)根
解：在方程?$2x^2+2x + 3 = 0$?中
?$a = 2，$??$b = 2，$??$c = 3$?
?$ ?=2^2-4×2×3=-20<0，$?
所以方程沒(méi)有實(shí)數(shù)根
解：在方程?$t^2-t + 3 = 0$?中
?$a = 1，$??$b=-1，$??$c = 3，$?
?$ ?=(-1)^2-4×1×3=-11<0$?
所以方程沒(méi)有實(shí)數(shù)根
解：在方程?$3x^2+8x = 0$?中
?$a = 3，$??$b = 8，$??$c = 0$?
?$ ?=8^2-4×3×0=64>0$?
所以方程有兩個(gè)不相等的實(shí)數(shù)根
解：將方程?$y + 1 = 3y^2$?
化為一般形式?$3y^2-y - 1 = 0$?
?$ $?其中?$a = 3，$??$b=-1，$??$c = -1$?
?$ ?=(-1)^2-4×3×(-1)=13>0$?
所以方程有兩個(gè)不相等的實(shí)數(shù)根
解：將方程?$4x(x - 1)=-1$?
化為一般形式?$4x^2-4x + 1 = 0$?
?$ $?其中?$a = 4，$??$b=-4，$??$c = 1$?
?$ ?=(-4)^2-4×4×1=0$?
所以方程有兩個(gè)相等的實(shí)數(shù)根
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