Proving a quadratic polynomial has no real roots without using derivatives or any formulas

Question

How do I prove that a quadratic polynomial has no real root without the use of any derivatives or formulas? The specific equation is $$x^2-6x+10=0$$

Answer 1

$x^2-6x+10=x^2-6x+9+1=(x-3)^2+1>0$ for all real numbers $x$.

Answer 2

Hint: By completing the square.

Answer 3

Without formulas:

The equation is that of a parabola with a vertical axis and the vertex is at coordinates obtained by completing the square. After using this transformation of the expression, you will notice that it is strictly positive.

Answer 4

Here's one way without completing the square.

Subtracting $10$ from both sides gives $x^2 - 6x = -10,$ which can be rewritten as $x(x-6) = -10.$ Now use the variable change $y = x-3$ (note that $x-3$ is the midpoint of $x$ and $x-6)$ to symmetrize the equation, getting $(y+3)(y-3) = -10.$ Expanding, we get $y^2 - 9 = -10,$ and adding $10$ to both sides gives $y^2 = -1,$ which has no real solution. (Note that if there had been a real solution for $x,$ say $x = r,$ then $y=r+3$ would be a real solution for $y.)$