Помогите решить:1)x^2-16x+28=0 2)x^2-27x=0 3)x^2-27x=0 4)3x^2-6x-7=0 5)5y^2+y-3=0

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