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Solution
I.$$6x^{2} + 29x + 35 = 0$$
=> $$6x^2 + 15x + 14x + 35 = 0$$
=> $$3x (2x + 5) + 7 (2x + 5) = 0$$
=> $$(2x + 5) (3x + 7) = 0$$
=> $$x = \frac{-7}{3} , \frac{-5}{2}$$
II.$$3y^{2} + 11y + 10 = 0$$
=> $$3y^2 + 6y + 5y + 10 = 0$$
=> $$3y (y + 2) + 5 (y + 2) = 0$$
=> $$(y + 2) (3y + 5) = 0$$
=> $$y = -2 , \frac{-5}{3}$$
Hence $$x < y$$
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