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# i don't get it!!!!!!!

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solve ax2+bx+c=0

Guest Feb 21, 2017
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### 1+0 Answers

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ax^2+bx+c=0      subtract c from both sides

ax^2 + bx  =  -c     divide through by a

x^2  + (b/a)x  =   -c/a    take  (1/2) of  (a/b)  =  b / [2a].....square this = b^2/[4a^2]...

add it  to both sides

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

Factor the left side.....get a common denominator on the right

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

Take  both square roots

x  + b/[2a]   =  ±  √  ( [ b^2 - 4ac] / [4a^2]  )

x  + b/[2a]   =  ±  √  ( [ b^2 - 4ac] / [2a]  )

Subtract  b/[2a]  from both sides

x  =  ±  √  ( [ b^2 - 4ac] / [2a]  )   -  b/[2a]

x =  ( -b  ±  √  ( [ b^2 - 4ac]  )  / [ 2a ]

Voila!!!!!......this is  the  Quadratic Formula.....

CPhill  Feb 21, 2017

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