(u+2)*(2u+3)=0

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

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

We multiply parentheses ..

(+2u^2+3u+4u+6)=0

We get rid of parentheses

2u^2+3u+4u+6=0

We add all the numbers together, and all the variables

2u^2+7u+6=0

a = 2; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·2·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*2}=\frac{-8}{4}
=-2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*2}=\frac{-6}{4}
=-1+1/2
$