(5+u)(3-u)=0

Solution for (5+u)(3-u)=0 equation:

(5+u)(3-u)=0

We add all the numbers together, and all the variables

(u+5)(-1u+3)=0

We multiply parentheses ..

(-1u^2+3u-5u+15)=0

We get rid of parentheses

-1u^2+3u-5u+15=0

We add all the numbers together, and all the variables

-1u^2-2u+15=0

a = -1; b = -2; c = +15;
Δ = b2-4ac
Δ = -22-4·(-1)·15
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*-1}=\frac{-6}{-2}
=+3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*-1}=\frac{10}{-2}
=-5
$