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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

4u^2+32u-9u-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
$