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

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Solution for (1+u)(3u-2)=0 equation:



(1+u)(3u-2)=0
We add all the numbers together, and all the variables
(u+1)(3u-2)=0
We multiply parentheses ..
(+3u^2-2u+3u-2)=0
We get rid of parentheses
3u^2-2u+3u-2=0
We add all the numbers together, and all the variables
3u^2+u-2=0
a = 3; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·3·(-2)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*3}=\frac{-6}{6} =-1 $
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*3}=\frac{4}{6} =2/3 $

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