-4t^2+32t=-132

解答

- 4 t^{2} + 32 t = - 132

t = - 3, t = 11

求解步骤

- 4 t^{2} + 32 t = - 132

将

132

para o lado esquerdo

- 4 t^{2} + 32 t + 132 = 0

使用求根公式求解

t_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 4 ) \cdot 132}{2 ( - 4 )}

3 2^{2} - 4 (- 4) \cdot 132 = 56

t_{1, 2} = \frac{- 32 \pm 56}{2 ( - 4 )}

将解分隔开

t_{1} = \frac{- 32 + 56}{2 ( - 4 )}, t_{2} = \frac{- 32 - 56}{2 ( - 4 )}

t = \frac{- 32 + 56}{2 ( - 4 )} : - 3

t = \frac{- 32 - 56}{2 ( - 4 )} : 11

二次方程组的解是:

t = - 3, t = 11

f a c t or 27 x^{2} + 5 = 48 x

f a c t or 4 y^{2} - 12 y + 9 = 0

- 2 t^{2} + 7 t - 4 = - 3 t^{2}

4 x^{2} - 80 x + 392 = 0

f a c t or x^{2} + 2 x - 4 = - 4