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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3u^2-8u+12u-32)=0

We get rid of parentheses

3u^2-8u+12u-32=0

We add all the numbers together, and all the variables

3u^2+4u-32=0

a = 3; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·3·(-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{-(4)-20}{2*3}=\frac{-24}{6}
=-4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*3}=\frac{16}{6}
=2+2/3
$