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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4u^2+36u-8u-72)=0

We get rid of parentheses

4u^2+36u-8u-72=0

We add all the numbers together, and all the variables

4u^2+28u-72=0

a = 4; b = 28; c = -72;
Δ = b2-4ac
Δ = 282-4·4·(-72)
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{-(28)-44}{2*4}=\frac{-72}{8}
=-9
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+44}{2*4}=\frac{16}{8}
=2
$