Question

12x^2+16x-3=0 Помогите решить пж

Answer 1

$\displaystyle 12x^2+16x-3=0\\12x^2+18x-2x-3=0\\6x(2x+3)-(2x+3)=0\\92x+3)(6x-1)=0\\2x+3=0,6x-1=0\\\\ x_{1}=-\frac{3}{2}=-1,5;x_{2}=\frac{1}{6}$