(3u+5)(8+u)=0

Solution for (3u+5)(8+u)=0 equation:

(3u+5)(8+u)=0

We add all the numbers together, and all the variables

(3u+5)(u+8)=0

We multiply parentheses ..

(+3u^2+24u+5u+40)=0

We get rid of parentheses

3u^2+24u+5u+40=0

We add all the numbers together, and all the variables

3u^2+29u+40=0

a = 3; b = 29; c = +40;
Δ = b2-4ac
Δ = 292-4·3·40
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-19}{2*3}=\frac{-48}{6}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+19}{2*3}=\frac{-10}{6}
=-1+2/3
$