(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+17u-18=0

a = -4; b = 17; c = -18;
Δ = b2-4ac
Δ = 172-4·(-4)·(-18)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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