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

Solution for (6-u)(4u-9)=0 equation:

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

We add all the numbers together, and all the variables

(-1u+6)(4u-9)=0

We multiply parentheses ..

(-4u^2+9u+24u-54)=0

We get rid of parentheses

-4u^2+9u+24u-54=0

We add all the numbers together, and all the variables

-4u^2+33u-54=0

a = -4; b = 33; c = -54;
Δ = b2-4ac
Δ = 332-4·(-4)·(-54)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-15}{2*-4}=\frac{-48}{-8}
=+6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+15}{2*-4}=\frac{-18}{-8}
=2+1/4
$