Ответы и объяснения

2013-10-07T17:49:50+00:00
1) x^2-16x+28=0
D=в^2-4ac=16^2-4*1*28=256-112=144
 x_{1}  = \frac{-b- \sqrt{D} }{2a} = \frac{-(-16)- \sqrt{144} }{2*1} =  \frac{16-12}{2} =2
 x_{2}= \frac{-b+ \sqrt{D} }{2a}  =  \frac{-(-16)+ \sqrt{144} }{2*1}=  \frac{16+12}{2} =14
4) 5y^2 + y - 3 = 0
5y^2 + y = 3
y (5y) = 3
y = 3 5y = 3 y = 3 - 5 y = - 2