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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3u^2-5u+9u+15)=0

We get rid of parentheses

-3u^2-5u+9u+15=0

We add all the numbers together, and all the variables

-3u^2+4u+15=0

a = -3; b = 4; c = +15;
Δ = b2-4ac
Δ = 42-4·(-3)·15
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*-3}=\frac{-18}{-6}
=+3
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*-3}=\frac{10}{-6}
=-1+2/3
$