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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4u^2+11u+45=0

a = -4; b = 11; c = +45;
Δ = b2-4ac
Δ = 112-4·(-4)·45
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-29}{2*-4}=\frac{-40}{-8}
=+5
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+29}{2*-4}=\frac{18}{-8}
=-2+1/4
$