(1+u)(4u-2)=0

Solution for (1+u)(4u-2)=0 equation:

(1+u)(4u-2)=0

We add all the numbers together, and all the variables

(u+1)(4u-2)=0

We multiply parentheses ..

(+4u^2-2u+4u-2)=0

We get rid of parentheses

4u^2-2u+4u-2=0

We add all the numbers together, and all the variables

4u^2+2u-2=0

a = 4; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·4·(-2)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*4}=\frac{-8}{8}
=-1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*4}=\frac{4}{8}
=1/2
$