(4u+2)(9-u)=0

Solution for (4u+2)(9-u)=0 equation:

(4u+2)(9-u)=0

We add all the numbers together, and all the variables

(4u+2)(-1u+9)=0

We multiply parentheses ..

(-4u^2+36u-2u+18)=0

We get rid of parentheses

-4u^2+36u-2u+18=0

We add all the numbers together, and all the variables

-4u^2+34u+18=0

a = -4; b = 34; c = +18;
Δ = b2-4ac
Δ = 342-4·(-4)·18
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-38}{2*-4}=\frac{-72}{-8}
=+9
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+38}{2*-4}=\frac{4}{-8}
=-1/2
$