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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4u^2-9u+8u+18)=0

We get rid of parentheses

-4u^2-9u+8u+18=0

We add all the numbers together, and all the variables

-4u^2-1u+18=0

a = -4; b = -1; c = +18;
Δ = b2-4ac
Δ = -12-4·(-4)·18
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*-4}=\frac{-16}{-8}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*-4}=\frac{18}{-8}
=-2+1/4
$