解の公式の求め方

a x 2 +bx+c=0 a{ x 2 + b a x+ ( b 2a ) 2 }−a ( b 2a ) 2 +c=0 a{ x 2 + b a x+ ( b 2a ) 2 }=a ( b 2a ) 2 −c { x 2 + b a x+ ( b 2a ) 2 }= ( b 2a ) 2 − c a ( x+ b 2a ) 2 = b 2 −4ac 4 a 2 x+ b 2a =± b 2 −4ac 4 a 2 x=− b 2a ± b 2 −4ac 2a x= −b± b 2 −4ac 2a

a x 2 +2bx+c=0 a{ x 2 +2b a x+ ( b a ) 2 }−a ( b a ) 2 +c=0 a{ x 2 + 2b a x+ ( b a ) 2 }=a ( b a ) 2 −c { x 2 + 2b a x+ ( b a ) 2 }= ( b a ) 2 − c a ( x+ b a ) 2 = b 2 −ac a 2 x+ b a =± b 2 −ac a 2 x=− b a ± b 2 −ac a x= −b± b 2 −ac a