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

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