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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4u^2+3u+28u-21)=0

We get rid of parentheses

-4u^2+3u+28u-21=0

We add all the numbers together, and all the variables

-4u^2+31u-21=0

a = -4; b = 31; c = -21;
Δ = b2-4ac
Δ = 312-4·(-4)·(-21)
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{-(31)-25}{2*-4}=\frac{-56}{-8}
=+7
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+25}{2*-4}=\frac{-6}{-8}
=3/4
$