помогите решить уравнение! 3х2-10х+7=0 2) -4Х2-12Х+7=0

3х²-10х+7=0
а=3, в=-10, с=7
Д=в²-4ас=-10²-4*3*7=100-84=16
х1= [tex] \frac{-b- \sqrt{D} }{2a} = \frac{10- \sqrt{16} }{2*3} = \frac{6}{3} = 2[/tex]
x2= [tex] \frac{-b+ \sqrt{D} }{2a} = \frac{10+ \sqrt{16} }{2*3} = \frac{14}{6} = \frac{7}{3} =2 \frac{1}{3} [/tex]
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-4x²-12х+7=0
а=-4, в=-12, с=7
Д=в²-4ас=-12²-4*(-4)*7=144+112=256
х1= [tex] \frac{-b- \sqrt{D} }{2a} = \frac{12- \sqrt{256} }{2*(-4)} = \frac{-4}{-8} = 0,5[/tex]
х2=  [tex] \frac{-b+ \sqrt{D} }{2a} = \frac{12+ \sqrt{256} }{2*(-4)} = \frac{28}{-8} = -3,5[/tex]