найдите корни уравнения (5y-7)(2y-0,4)

=10[tex] y^{2} [/tex]-2y-14y+2,8=10[tex] y^{2} [/tex]-16y+2,8=5[tex] y^{2} [/tex]-8y+1.4
D=[tex] (-8)^{2} [/tex]-4*5*1,4=64-28=36
x1,2=[tex] \frac{8+- \sqrt{36} }{2*5} = \frac{8+-6}{10} [/tex]
x1=[tex] \frac{8-6}{10} = \frac{2}{10} =0,2[/tex]
x2=[tex] \frac{8+6}{10} = \frac{14}{10} =1,4[/tex]