(2-u)(4u+6)=0

Solution for (2-u)(4u+6)=0 equation:

(2-u)(4u+6)=0

We add all the numbers together, and all the variables

(-1u+2)(4u+6)=0

We multiply parentheses ..

(-4u^2-6u+8u+12)=0

We get rid of parentheses

-4u^2-6u+8u+12=0

We add all the numbers together, and all the variables

-4u^2+2u+12=0

a = -4; b = 2; c = +12;
Δ = b2-4ac
Δ = 22-4·(-4)·12
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*-4}=\frac{-16}{-8}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*-4}=\frac{12}{-8}
=-1+1/2
$