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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4u^2+5u+16u-20)=0

We get rid of parentheses

-4u^2+5u+16u-20=0

We add all the numbers together, and all the variables

-4u^2+21u-20=0

a = -4; b = 21; c = -20;
Δ = b2-4ac
Δ = 212-4·(-4)·(-20)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*-4}=\frac{-32}{-8}
=+4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*-4}=\frac{-10}{-8}
=1+1/4
$