^{2}+bx+c=0 Multiply by 4a to get

^{2}x

^{2}+4abx+4ac=0 Compete the square as

^{2}=4a

^{2}x

^{2}+4abx+b

^{2}so that

^{2}x

^{2}+4abx+4ac=(2ax+b)

^{2}-b

^{2}+4ac=0

^{2}=b

^{2}-4ac

^{2}-4ac)

^{2}-4ac)

^{2}-4ac))/(2a)

how do we solve it

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This answer comes from Littlewood's A University Algebra

ax^{2}+bx+c=0 Multiply by 4a to get

4a^{2}x^{2}+4abx+4ac=0 Compete the square as

(2ax+b)^{2}=4a^{2}x^{2}+4abx+b^{2} so that

4a^{2}x^{2}+4abx+4ac=(2ax+b)^{2}-b^{2}+4ac=0

(2ax+b)^{2}=b^{2}-4ac

2ax+b=±√(b^{2}-4ac)

2ax=-b±√(b^{2}-4ac)

x=(-b±√(b^{2}-4ac))/(2a)

Quadratic equation is derived by completing the square:

ax^2 + bx + c = 0

Subtract c from both sides:

ax^2 + bx = -c

Divide by a:

x^2 + (b/a)x = -c/a

Complete the square by taking half of b/a and squaring it, then adding to both sides:

Note: (b/a) * 1/2 = b/(2a)

x^2 + (b/a)x + (b/(2a))^2 = -c/a + (b/(2a))^2

Simplify:

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

Factor:

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

Take the square root:

x + (b/(2a)) = +/- sqrt(-c/a + b^2/(4a^2))

Subtract b/(2a) from both sides:

x = -b/(2a) +/- sqrt(-c/a + b^2/(4a^2))

Simplify:

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

------- + ---------

4a^2 4a^2

x = -b +/- sqrt (b^2 - 4ac)

--------------------------

2a

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