$-\frac{5}{2}\pm\frac{\sqrt{29}}{2}$
$解:3x^{2}+10x+15=0 $ $a=3,b=10,c=15$ $△=100-4×3×15=-80\lt 0$ $∴原方程无实数根$
$解:解答有错误。正确的解答过程：原方程化为x^{2}+3x−2=0。$ $\because a = 1,b = 3,c = -2，$$\therefore b^{2}-4ac = 9 + 8 = 17。$$\therefore x=\frac{-3\pm\sqrt{17}}{2}，$即$x_1=\frac{-3+\sqrt{17}}{2},x_2=\frac{-3-\sqrt{17}}{2}$
$解: 由题意，得[−(3m−1)]^{2}−4m(2m−1)=1，即m^{2}−2m=0，解得m_{1}=0,m_{2}=2。易知m≠0，∴m=2。∴原方程为2x^{2}−5x+3=0，解得x_{1}=1,x_{2}=\frac {3}{2}$
$解:x^{2}-x-3=0$ $a=1,b=-1,c=-3$ $△=1+12=13$ $x=\frac {1±\sqrt {13}}{2}$ $x_{1}=\frac {1+\sqrt {13}}{2},x_{2}=\frac {1-\sqrt {13}}{2}$
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