Quadratic equation

The given equation is the same as $ax^2+(a+b-c)x+b=0$ Then we consider the $D=(a+b-c)^2-4ab$ $D=(a+b-c)^2-(2\sqrt{ab})^2$ $=(a+2\sqrt{ab}+b-c)(a-2\sqrt{ab}+b-c)$ $=[(\sqrt{a}+\sqrt{b})^2-(\sqrt{c})^2][(\sqrt{a}-\sqrt{b})^2-(\sqrt{c})^2]$ $=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}-\sqrt{c})$ Since $\sqrt{a},\sqrt{b},\sqrt{c}$ are three sides of a triangle, we have the following inequalities: $\sqrt{a}+\sqrt{b}+\sqrt{c}>0$
$\sqrt{a}+\sqrt{b}-\sqrt{c}>0$ $\sqrt{a}-\sqrt{b}+\sqrt{c}>0$
and $\sqrt{a}-\sqrt{b}-\sqrt{c}<0$ So $D<0$ Hence, the equation has no real solution.

To make things look little easier, I will rewrite the problem so that the sides of the triangle are $a, b, c$ and the equation is

$a^2x^2+(a^2+b^2-c^2)x=-b^2$

From the law of cosines $c^2=a^2+b^2-2ab\cos C$

So that the equation becomes $a^2x^2+2abx\cos C+b^2=0$

For $C$ to be the angle at vertex $C$ of a non-degenerate triangle, $|\cos C|<1$ , so the equation does not have a solution.

The first numbers that would come into the mind will be $3,4,5$ taking the values of $\sqrt{a},\sqrt{b},\sqrt{c}$ respectively, whose squares are $9,16,25$ .

Substituting in the equation, we get $9x^2=-16$

Clearly indicating that there is no real solution existing.