Find number of solutions to $x^2-6x+\left[x\right]+7=0$

Question

Find number of solutions to $$x^2-6x+\lfloor x\rfloor+7=0$$

My Try: I have written given equation as $$(x-3)^2+\lfloor x \rfloor -2=0$$

$\implies$

$$(x-3)^2+\lfloor x-3 \rfloor +1=0$$

Letting $x-3=t$ we have

$$t^2+\lfloor t \rfloor +1=0$$

Obviously $t \ngtr0$ and $t \notin \mathbb{Z}$

So

$t \lt 0$

Any clue to proceed further?

Answer 1

Hint: $t^2+1\geqslant 2|t|\geqslant -\lfloor t \rfloor$, with equality never possible.

Answer 2

Since $[x]=x-f$ for some f in [0,1), you have $x^2-5x+7-f=0$. Pretending f is just a parameter for a moment, find solutions using the quadratic formula. Then vary f until those solutions are integers (If possible).