ax^2+bx+c=0,x

解答

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

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

求解步骤

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

使用求根公式求解

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

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

x = \frac{- b - b ^{2} - 4 a c}{2 a}; a \neq = 0

二次方程组的解是:

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

9 q^{2} - 387 q + 3510 = 0

8 a^{2} + 28 a = - 20

8 x^{2} - 19 = 15

3 x^{2} - 2.2 + 5 x = 0

1000000 = 25 t^{2} + 72 t + 100