Answer to Question #12158 in Algebra for hailey
x1 = [ -b + √(b²-4c) ] / 2
x2 = [ -b - √(b²-4c) ] / 2.
So, we can factor it as
x² + bx + c = (x-x1)(x-x2).
In a case of trinomial ax² + bx + c we can factorize it as
ax² + bx + c = a(x² + b/a·x + c/a),
where the trinomial x² + b/a·x + c/a has the same form as x² + bx + c.
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