(2u+9)(5-u)=0

Solution for (2u+9)(5-u)=0 equation:

(2u+9)(5-u)=0

We add all the numbers together, and all the variables

(2u+9)(-1u+5)=0

We multiply parentheses ..

(-2u^2+10u-9u+45)=0

We get rid of parentheses

-2u^2+10u-9u+45=0

We add all the numbers together, and all the variables

-2u^2+u+45=0

a = -2; b = 1; c = +45;
Δ = b2-4ac
Δ = 12-4·(-2)·45
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{-(1)-19}{2*-2}=\frac{-20}{-4}
=+5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*-2}=\frac{18}{-4}
=-4+1/2
$