if (x2 -5x+3)(y2 +y+1)<2y for all y belongs to the set of real numbers,then set of all values of x is

Since y²+y+1 is always positive, we can divide by this quantity on both sides to get the inequality

x² - 5x + 3 < 2y/(y²+y+1) or equivalently x² - 5x + 3 < 2/(y+1/y+1)

When y is non-negative, min val of RHS is zero.

If y<0 then y + 1/y <=-2 so, y+1/y+1 <=-1 and hence RHS >=-2.

Hence we need x² - 5x + 3<-2 or x² - 5x + 5<0 which is easy to solve