(2u+1)(9-u)=0

Solution for (2u+1)(9-u)=0 equation:

(2u+1)(9-u)=0

We add all the numbers together, and all the variables

(2u+1)(-1u+9)=0

We multiply parentheses ..

(-2u^2+18u-1u+9)=0

We get rid of parentheses

-2u^2+18u-1u+9=0

We add all the numbers together, and all the variables

-2u^2+17u+9=0

a = -2; b = 17; c = +9;
Δ = b2-4ac
Δ = 172-4·(-2)·9
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*-2}=\frac{-36}{-4}
=+9
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*-2}=\frac{2}{-4}
=-1/2
$