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

Solution for (6-u)(3u+5)=0 equation:

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

We add all the numbers together, and all the variables

(-1u+6)(3u+5)=0

We multiply parentheses ..

(-3u^2-5u+18u+30)=0

We get rid of parentheses

-3u^2-5u+18u+30=0

We add all the numbers together, and all the variables

-3u^2+13u+30=0

a = -3; b = 13; c = +30;
Δ = b2-4ac
Δ = 132-4·(-3)·30
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-23}{2*-3}=\frac{-36}{-6}
=+6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+23}{2*-3}=\frac{10}{-6}
=-1+2/3
$