If ax*x + bx +c = 0, then what is x

To solve ax*x+bx+c =
0.

Solution: We can write the equation
like:

ax^2+bx+c= 0.....(1), as x*x = x^2. This is a second
degree equation and is also called a quadratic equation which will be solved by writing
the left as diffrence of an unknown square and a known square equal to
zero:

LHS of (1) :

ax^2+bx+c =
a[x^2+(b/a)x]+c

=a(x^2+(b/a)x+(b/(2a))^2] - b^2/(4a)+c.
Here (b^2/(4a) is added and subtracted.

=a[x+(b/(2a))]^2 -
(b^2-4ac)/4a .

=a{[x+b/(2a)]^2 -
(b^2-4ac)/(2a)^2}

So the equation (1) now could be
rewritten like:

a{[x+b/(2a)]^2 - (b^2-4ac)/(2a)^2} = 0. Or
bt dividing by a which is not equal to zero, we
get:

[x+b/(2a)]^2  - (b^2-4ac)/(2a)^2 = 0.
Or

[x+b/(2a)]^2 = (b^-4ac)/(2a)^2. Taking the square
root,

x+b/(2a) = +sqrt(b^2-4ac)/2a . Or x+b/(2a) =
-sqrt(b^2-4ac)/2a. So we get 2 solutions from the se two
results.

x1 = [-b+sqrt(b^2-4ac)]/(2a)
Or

x2 = [-b-sqrt(b^2-4ac)]/(2a)