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

Solution for (2u+3)(6+u)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2u^2+12u+3u+18)=0

We get rid of parentheses

2u^2+12u+3u+18=0

We add all the numbers together, and all the variables

2u^2+15u+18=0

a = 2; b = 15; c = +18;
Δ = b2-4ac
Δ = 152-4·2·18
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-9}{2*2}=\frac{-24}{4}
=-6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+9}{2*2}=\frac{-6}{4}
=-1+1/2
$