Answer:
[tex]x = (-b + \sqrt{b^{2} - 4\, a\, c}) / (2\, a)[/tex] or [tex]x = (-b - \sqrt{b^{2} - 4\, a\, c}) / (2\, a)[/tex] assuming that [tex]b^{2} - 4\, a\, c \ge 0[/tex].
Step-by-step explanation:
Let [tex]h[/tex] and [tex]k[/tex] be constants. Consider [tex]a\, (x + h)^{2} = k[/tex]. In this equation, [tex](x + h)^{2}[/tex], the only term that includes [tex]x[/tex], is a perfect square. If [tex]k \ge 0[/tex], solving this equation is as simple as taking the square root of both sides of the equation:
[tex]x + h = \sqrt{k / a}[/tex] or [tex]x + h = -\sqrt{k / a}[/tex].
[tex]x = (-h) + \sqrt{k / a}[/tex] or [tex]x = (-h) - \sqrt{k / a}[/tex].
Assume that there are values for [tex]h[/tex] and [tex]k[/tex] such that [tex]a\, x^{2} + b\, x + c = 0[/tex] is equivalent to [tex]a\, (x + h)^{2} = k[/tex]. If [tex](k / a) \ge 0[/tex], then [tex]x = (-h) + \sqrt{k / a}[/tex] and [tex]x = (-h) - \sqrt{k / a}[/tex] would be solutions to [tex]a\, x^{2} + b\, x + c = 0\![/tex].
Apply binomial expansion to [tex]a\, (x + h)^{2} = k[/tex] and rewrite to find the values for [tex]h[/tex] and [tex]k[/tex]:
[tex]a\, (x^{2} + 2\, h\, x + h^{2}) - k = 0[/tex].
[tex]a\, x^{2} + 2\, a\, h\, x + (a\, h^{2} - k) = 0[/tex].
Match the coefficients of this equation with those in [tex]a\, x^{2} + b\, x + c = 0[/tex]:
[tex]2\, a\, h = b[/tex].
[tex]a\, h^{2} - k = c[/tex].
Solve for [tex]h[/tex] and [tex]k[/tex] in terms of [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex]:
[tex]h = (b / 2\, a)[/tex].
[tex]\begin{aligned}k &= a\, h^{2} - c \\ &= \frac{a\, b^{2}}{4\, a^{2}} - c \\ &= \frac{b^{2}}{4\, a} - c \\ &= \frac{b^{2} - 4\, a\, c}{4\, a}\end{aligned}[/tex].
Hence, as long as [tex](b^{2} - 4\, a\, c) \ge 0[/tex], (such that [tex](k / a) \ge 0[/tex],) solutions to [tex]a\, x^{2} + b\, x + c = 0[/tex] would be:
[tex]\begin{aligned}x &= (-h) + \sqrt{\frac{k}{a}} \\ &= -\frac{b}{2\, a} + \sqrt{\frac{b^{2} - 4\, a\, c}{4\, a^{2}}} \\ &= -\frac{b}{2\, a} + \frac{\sqrt{b^{2} - 4\, a\, c}}{2\, a}\end{aligned}[/tex], and
[tex]\begin{aligned}x &= (-h) - \sqrt{\frac{k}{a}} \\ &= -\frac{b}{2\, a} - \sqrt{\frac{b^{2} - 4\, a\, c}{4\, a^{2}}} \\ &= -\frac{b}{2\, a} - \frac{\sqrt{b^{2} - 4\, a\, c}}{2\, a}\end{aligned}[/tex].
