$(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions iff $7a^2 - 50a + 7\leq 0 $ How can we show that $(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions if and only if $7a^2 - 50a + 7\leq 0$?
I know these are quadratics and can solve them, but I'm not entirely sure what the question is asking of me and how to lay out the logic.

From the quadratic formula you know that the solutions of
$$(a-1)x^2+3(a+1)x+4(a-1)=0$$
are given by
$$x=\frac{-3(a+1)\pm\sqrt{9(a+1)^2-16(a-1)^2}}{2(a-1)}\;.$$
These will be real if and only if
$$9(a+1)^2-16(a-1)^2\ge 0\;.$$
Expanding the lefthand side, we see that this inequality reduces to 
$$-7a^2+50a-7\ge 0\;.$$
Now just multiply the inequality by $-1$.
