$$ a B+x^{2}+m x+a=0 $$ बरा $$ x $$ $$ c+a=a b $$ $$ 2 a+c $$. अथबा $$ (2 a+c)^{2}=2 b^{2} $$

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We begin with the equation
$$
aB + x^2 + mx + a = 0.
$$
Since the letter $$B$$ later appears in a relation with $$b$$ (namely, in the equation $$c+a=ab$$) it is natural to assume that
$$
aB = ab,
$$
so that the equation may be written as
$$
x^2 + mx + a(b+1) = 0.
$$

Once written in this form the solution in $$x$$ is given by the quadratic‐formula:
$$
x=\frac{-m\pm\sqrt{m^2-4a(b+1)}}{2}.
$$

There is an additional relation given:
$$
c+a=ab.
$$
Subtracting $$a$$ from both sides we find
$$
c=a(b-1).
$$

In view of this relation we now compute
$$
2a+c=2a+a(b-1)=a(b+1).
$$
Also the problem gives the alternative condition
$$
(2a+c)^2=2b^2.
$$
Since
$$
2a+c=a(b+1),
$$
this condition becomes
$$
\bigl(a(b+1)\bigr)^2=2b^2,
$$
or
$$
a^2(b+1)^2=2b^2.
$$
Solving for $$a$$ one obtains
$$
a=\pm\sqrt{\frac{2b^2}{(b+1)^2}}=\pm\frac{\sqrt{2}\,|b|}{b+1}.
$$

Thus, the answer we have found is that the solution of
$$
aB+x^2+mx+a=0
$$
(with $$aB=ab$$ and the relation $$c+a=ab$$) is
$$
\boxed{x=\frac{-m\pm\sqrt{m^2-4a(b+1)}}{2}},
$$
where we note that
$$
2a+c=a(b+1)
$$
and the alternative condition is
$$
(2a+c)^2=2b^2.
$$
The solution to the equation $$ aB + x^2 + mx + a = 0 $$ is: $$ x = \frac{-m \pm \sqrt{m^2 - 4a(b+1)}}{2} $$ where $$ a = \pm \frac{\sqrt{2} \, |b|}{b+1} $$.

\( a B+x^{2}+m x+a=0 \) बरा \( x \) \[ c+a=a b \] \( 2 a+c \). अथबा \( (2 a+c)^{2}=2 b^{2} \)

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