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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4u^2+36u+5u-45)=0

We get rid of parentheses

-4u^2+36u+5u-45=0

We add all the numbers together, and all the variables

-4u^2+41u-45=0

a = -4; b = 41; c = -45;
Δ = b2-4ac
Δ = 412-4·(-4)·(-45)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-31}{2*-4}=\frac{-72}{-8}
=+9
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+31}{2*-4}=\frac{-10}{-8}
=1+1/4
$