(3u+5)(u+1)=0

Solution for (3u+5)(u+1)=0 equation:

(3u+5)(u+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3u^2+8u+5=0

a = 3; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·3·5
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{-(8)-2}{2*3}=\frac{-10}{6}
=-1+2/3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*3}=\frac{-6}{6}
=-1
$