(2u-9)(8+u)=0

Solution for (2u-9)(8+u)=0 equation:

(2u-9)(8+u)=0

We add all the numbers together, and all the variables

(2u-9)(u+8)=0

We multiply parentheses ..

(+2u^2+16u-9u-72)=0

We get rid of parentheses

2u^2+16u-9u-72=0

We add all the numbers together, and all the variables

2u^2+7u-72=0

a = 2; b = 7; c = -72;
Δ = b2-4ac
Δ = 72-4·2·(-72)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-25}{2*2}=\frac{-32}{4}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+25}{2*2}=\frac{18}{4}
=4+1/2
$