Решить уравнения : а)2*(5х+7)*(1+х)=0 б)3х в квадрате + 2х-5=0

[latex]1)2(5x+7)(1+x)=0\\(10x+14)(1+x)=0\\10x+14=0\\10x=-14\\x=- \frac{14}{10} \\ x=-1.4 \\ \\ 1+x=0\\x=-1 \\ \\ \\ [/latex] [latex]3 x^{2} +2x-5=0\\D=4+60=64\\ \sqrt{D} =8 \\ x_{1} = \frac{-2+8}{6} = \frac{6}{6} =1 \\ \\ x_{2} = \frac{-2-8}{6} =- \frac{10}{6} =- \frac{5}{3} =-1 \frac{2}{3} [/latex]