(8-u)(4u+9)=0

Solution for (8-u)(4u+9)=0 equation:

(8-u)(4u+9)=0

We add all the numbers together, and all the variables

(-1u+8)(4u+9)=0

We multiply parentheses ..

(-4u^2-9u+32u+72)=0

We get rid of parentheses

-4u^2-9u+32u+72=0

We add all the numbers together, and all the variables

-4u^2+23u+72=0

a = -4; b = 23; c = +72;
Δ = b2-4ac
Δ = 232-4·(-4)·72
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-41}{2*-4}=\frac{-64}{-8}
=+8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+41}{2*-4}=\frac{18}{-8}
=-2+1/4
$