Proving a relation related to quadratic equation

Instead doing all the hard work you did, you can notice that the difference of roots $(\vert x_1-x_2\vert )$ is same for both the equations. Hence :

$$|\alpha-\beta|=|(\alpha+\delta)-(\beta+\delta)|=\sqrt{(\alpha+\beta)^2-4\alpha \beta}=\sqrt{(\alpha+\delta +\beta+ \delta)^2-4(\alpha+\delta)( \beta+\delta)}$$

$$\implies \sqrt{\left(\frac {-2b}{a} \right)^2 -4\left(\frac ca \right)}= \sqrt{\left(\frac {-2B}{A} \right)^2 -4\left(\frac CA\right)}$$

$$\implies \frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$$

Alternative hint: we can assume WLOG that $\,a=A=1\,$, since both the roots and the equality to be proved are homogeneous in the respective coefficients.

Then, if $\alpha, \beta$ are the roots of $x^2+2bx+c=0$, the polynomial with roots $\alpha+\delta, \beta+\delta$ is:

$$(x-\delta)^2+2b (x-\delta)+c=0 \;\;\iff\;\; x^2 + 2(b-\delta)x+\delta^2-2b\delta+c=0\,$$

Identifying coefficients gives $B=b-\delta$ and $C=\delta^2-2b\delta+c\,$, then:

$$\require{cancel} B^2 - C=(b^2-\bcancel{2 b\delta}+\cancel{\delta^2})-(\cancel{\delta^2}-\bcancel{2b\delta}+c) = b^2 - c $$

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[ EDIT ]  To answer OP's edit:

A continuation of my method would be more appreciated compared to other methods.

There is an error/typo in formula (2). Once corrected (in red below): $$\require{cancel} \frac{4b^2-2ac}{a^\bcancel{2}}=\frac{4aB^2-2A^2 c-4Aa\color{red}{C}+4cA^2}{A^2 \bcancel{a}} $$

$$ 4b^2A^2-\bcancel{2acA^2}=4a^2B^2-\bcancel{2acA^2}-4a^2AC+4acA^2 $$

$$ \bcancel{4}A^2(b^2-ac) = \bcancel{4}a^2(B^2-AC) $$

$$\frac{b^2-ac}{a^2} =\frac{B^2-AC}{A^2} $$

WLOG, $A=a=1$ (otherwise you can divide the trinomials by their leading coefficient).

Then

$$(x+\delta)^2+2B(x+\delta)+C=x^2+2(\delta+B)x+\delta^2+2B\delta+C=x^2+2bx+c,$$ and

$$b^2-c=(\delta+B)^2-(\delta^2+2B\delta+C)=B^2-C.$$