(4u-9)(5-u)=0

Solution for (4u-9)(5-u)=0 equation:

(4u-9)(5-u)=0

We add all the numbers together, and all the variables

(4u-9)(-1u+5)=0

We multiply parentheses ..

(-4u^2+20u+9u-45)=0

We get rid of parentheses

-4u^2+20u+9u-45=0

We add all the numbers together, and all the variables

-4u^2+29u-45=0

a = -4; b = 29; c = -45;
Δ = b2-4ac
Δ = 292-4·(-4)·(-45)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-11}{2*-4}=\frac{-40}{-8}
=+5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+11}{2*-4}=\frac{-18}{-8}
=2+1/4
$