(3u+5)(2+u)=0

Solution for (3u+5)(2+u)=0 equation:

(3u+5)(2+u)=0

We add all the numbers together, and all the variables

(3u+5)(u+2)=0

We multiply parentheses ..

(+3u^2+6u+5u+10)=0

We get rid of parentheses

3u^2+6u+5u+10=0

We add all the numbers together, and all the variables

3u^2+11u+10=0

a = 3; b = 11; c = +10;
Δ = b2-4ac
Δ = 112-4·3·10
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{-(11)-1}{2*3}=\frac{-12}{6}
=-2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*3}=\frac{-10}{6}
=-1+2/3
$