Best Answer

The essence of the proof is simply to complete the square for a generalised quadratic equation. Like this:

ax2 + bx + c = 0

Take 'a' outside:

a[x2 + bx/a + c/a] = 0

Divide through by 'a':

x2 + bx/a + c/a = 0

Complete the square:

(x + b/2a)2 - b2/4a2 + c/a = 0

Rearrange to find x:

(x + b/2a)2 = b2/4a2 - c/a

x + b/2a = (+/-)sqrt[b2/4a2 - c/a]

x = -b/2a (+/-) sqrt[b2/4a2 - c/a]

Finally, fiddle around so that (1/2a) can be taken out as a common factor:

x = -b/2a (+/-) sqrt[b2/4a2 - 4ac/4a2]

x = -b/2a (+/-) sqrt[(1/4a2)(b2 - 4ac)]

x = -b/2a (+/-) sqrt(1/4a2)sqrt(b2 - 4ac)

x = -b/2a (+/-) (1/2a)sqrt(b2 - 4ac)

x = [ -b (+/-) sqrt(b2 - 4ac) ] / 2a.

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