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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4u^2+2u+24u+12)=0

We get rid of parentheses

4u^2+2u+24u+12=0

We add all the numbers together, and all the variables

4u^2+26u+12=0

a = 4; b = 26; c = +12;
Δ = b2-4ac
Δ = 262-4·4·12
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*4}=\frac{-48}{8}
=-6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*4}=\frac{-4}{8}
=-1/2
$