How do you solve $(2 u + 7) (3 u - 1) = 0$ $(2u+7)(3u-1)=0$ ?

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How do you solve $(2 u + 7) (3 u - 1) = 0$ $(2u+7)(3u-1)=0$ ?

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Answer 1 · 2018-04-16T06:26:43

$u = \frac{- 7}{2}$ $u=(-7)/2$
$u = \frac{1}{3}$ $u=1/3$

Explanation:

Given

$(2 u + 7) (3 u - 1) = 0$ $(2u+7)(3u-1)=0$

$2 u + 7 = 0$ $2u+7=0$

$2 u = - 7$ $2u=-7$

$u = \frac{- 7}{2}$ $u=(-7)/2$

$(3 u - 1)$ $(3u-1)$

$3 u = 1$ $3u=1$

$u = \frac{1}{3}$ $u=1/3$

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How do you solve $(2 u + 7) (3 u - 1) = 0$ $(2u+7)(3u-1)=0$ ?

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How do you solve (2u+7)(3u-1)=0(2u+7)(3u−1)=0(2u+7)(3u-1)=0?

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How do you solve $(2 u + 7) (3 u - 1) = 0$ $(2u+7)(3u-1)=0$ ?