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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5u^2+16u-16=0

a = 5; b = 16; c = -16;
Δ = b2-4ac
Δ = 162-4·5·(-16)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*5}=\frac{-40}{10}
=-4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*5}=\frac{8}{10}
=4/5
$