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人気のある代数 >

(x)=ax^2+bx+c

解

(x) = a x^{2} + b x + c

解

x = \frac{- b + 1 + ( b - 1 ) ^{2} - 4 a c}{2 a}, x = \frac{- b + 1 - ( b - 1 ) ^{2} - 4 a c}{2 a}; a \neq = 0

解答ステップ

(x) = a x^{2} + b x + c

改良

x = a x^{2} + b x + c

辺を交換する

a x^{2} + b x + c = x

x

を左側に移動します

a x^{2} + b x + c - x = 0

標準的な形式で書く

a x^{2} + b x + c = 0

a x^{2} + (b - 1) x + c = 0

解くとthe二次式

x_{1, 2} = \frac{- ( b - 1 ) \pm ( b - 1 ) ^{2} - 4 a c}{2 a}

x = \frac{- ( b - 1 ) + ( b - 1 ) ^{2} - 4 a c}{2 a} : \frac{- b + 1 + ( b - 1 ) ^{2} - 4 a c}{2 a}

x = \frac{- ( b - 1 ) - ( b - 1 ) ^{2} - 4 a c}{2 a} : \frac{- b + 1 - ( b - 1 ) ^{2} - 4 a c}{2 a}

二次equationの解:

x = \frac{- b + 1 + ( b - 1 ) ^{2} - 4 a c}{2 a}, x = \frac{- b + 1 - ( b - 1 ) ^{2} - 4 a c}{2 a}; a \neq = 0

グラフ

人気の例

solvefor x,3x^2+4y^2+3x-2y+3=0

so l v e f or x, 3 x^{2} + 4 y^{2} + 3 x - 2 y + 3 = 0

3 x^{2} = 16

4 x^{2} = - 12 x + 4

(x^2+1)/2-(2x-3)/4+(x^2)/6 = 59/12

\frac{x ^{2} + 1}{2} - \frac{2 x - 3}{4} + \frac{x ^{2}}{6} = \frac{59}{12}

(x+1)^2-(x-2)^2=(x+3)^2+x^2-20

(x + 1)^{2} - (x - 2)^{2} = (x + 3)^{2} + x^{2} - 20