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what are the steps to solve this ax2 + bx + c = 0"

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2 Answers

This answer comes from Littlewood's A University Algebra
ax2+bx+c=0  Multiply by 4a to get
4a2x2+4abx+4ac=0 Compete the square as
(2ax+b)2=4a2x2+4abx+b2 so that
 
4a2x2+4abx+4ac=(2ax+b)2-b2+4ac=0
(2ax+b)2=b2-4ac
2ax+b=±√(b2-4ac)
2ax=-b±√(b2-4ac)
x=(-b±√(b2-4ac))/(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