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Solution
I.$$2x^{2} + 15x + 28 = 0$$
=> $$2x^2 + 8x + 7x + 28 = 0$$
=> $$2x (x + 4) + 7 (x + 4) = 0$$
=> $$(x + 4) (2x + 7) = 0$$
=> $$x = -4 , \frac{-7}{2}$$
II.$$4y^{2} + 18y + 14 = 0$$
=> $$4y^2 + 4y + 14y + 14 = 0$$
=> $$4y (y + 1) + 14 (y +1) = 0$$
=> $$(y + 1) (4y + 14) = 0$$
=> $$y = -1 , \frac{-7}{2}$$
$$\therefore x \leq y$$
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