(6-u)(3u+7)=0

Solution for (6-u)(3u+7)=0 equation:

(6-u)(3u+7)=0

We add all the numbers together, and all the variables

(-1u+6)(3u+7)=0

We multiply parentheses ..

(-3u^2-7u+18u+42)=0

We get rid of parentheses

-3u^2-7u+18u+42=0

We add all the numbers together, and all the variables

-3u^2+11u+42=0

a = -3; b = 11; c = +42;
Δ = b2-4ac
Δ = 112-4·(-3)·42
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-25}{2*-3}=\frac{-36}{-6}
=+6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+25}{2*-3}=\frac{14}{-6}
=-2+1/3
$