(4-u)(3u-7)=0

Solution for (4-u)(3u-7)=0 equation:

(4-u)(3u-7)=0

We add all the numbers together, and all the variables

(-1u+4)(3u-7)=0

We multiply parentheses ..

(-3u^2+7u+12u-28)=0

We get rid of parentheses

-3u^2+7u+12u-28=0

We add all the numbers together, and all the variables

-3u^2+19u-28=0

a = -3; b = 19; c = -28;
Δ = b2-4ac
Δ = 192-4·(-3)·(-28)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*-3}=\frac{-24}{-6}
=+4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*-3}=\frac{-14}{-6}
=2+1/3
$