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

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
$