(8+u)(5u-4)=0

Solution for (8+u)(5u-4)=0 equation:

(8+u)(5u-4)=0

We add all the numbers together, and all the variables

(u+8)(5u-4)=0

We multiply parentheses ..

(+5u^2-4u+40u-32)=0

We get rid of parentheses

5u^2-4u+40u-32=0

We add all the numbers together, and all the variables

5u^2+36u-32=0

a = 5; b = 36; c = -32;
Δ = b2-4ac
Δ = 362-4·5·(-32)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-44}{2*5}=\frac{-80}{10}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+44}{2*5}=\frac{8}{10}
=4/5
$