(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

(u+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+14u+12=0

a = 4; b = 14; c = +12;
Δ = b2-4ac
Δ = 142-4·4·12
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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