(8+u)(3u+4)=0

Solution for (8+u)(3u+4)=0 equation:

(8+u)(3u+4)=0

We add all the numbers together, and all the variables

(u+8)(3u+4)=0

We multiply parentheses ..

(+3u^2+4u+24u+32)=0

We get rid of parentheses

3u^2+4u+24u+32=0

We add all the numbers together, and all the variables

3u^2+28u+32=0

a = 3; b = 28; c = +32;
Δ = b2-4ac
Δ = 282-4·3·32
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20}{2*3}=\frac{-48}{6}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20}{2*3}=\frac{-8}{6}
=-1+1/3
$