(8-u)(2u+5)=0

Solution for (8-u)(2u+5)=0 equation:

(8-u)(2u+5)=0

We add all the numbers together, and all the variables

(-1u+8)(2u+5)=0

We multiply parentheses ..

(-2u^2-5u+16u+40)=0

We get rid of parentheses

-2u^2-5u+16u+40=0

We add all the numbers together, and all the variables

-2u^2+11u+40=0

a = -2; b = 11; c = +40;
Δ = b2-4ac
Δ = 112-4·(-2)·40
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*-2}=\frac{-32}{-4}
=+8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*-2}=\frac{10}{-4}
=-2+1/2
$