[tex]3y ^{2} + 6y -12x + 160=0 \\\\To f\ ind \ the \ x-intercept,\ set \ y = 0 \\\\3 \cdot 0 ^{2} + 6 \cdot 0 -12x + 160=0 \\ \\-12x+160=0 \ \ |\ subtract\ 2x\ to\ both\ sides \\ \\-12x=-160 \ \ | \ divide \ both \ sides\ by\ (-12 )\\\\x=\frac{160}{12} =\frac{40}{3}=13\frac{1}{3} [/tex]
