(2u-u)(4u+3)=0

Solution for (2u-u)(4u+3)=0 equation:

(2u-u)(4u+3)=0

We add all the numbers together, and all the variables

(+u)(4u+3)=0

We multiply parentheses ..

(+4u^2+3u)=0

We get rid of parentheses

4u^2+3u=0

a = 4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*4}=\frac{-6}{8}
=-3/4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*4}=\frac{0}{8}
=0
$