(7-u)(4u-6)=0

Solution for (7-u)(4u-6)=0 equation:

(7-u)(4u-6)=0

We add all the numbers together, and all the variables

(-1u+7)(4u-6)=0

We multiply parentheses ..

(-4u^2+6u+28u-42)=0

We get rid of parentheses

-4u^2+6u+28u-42=0

We add all the numbers together, and all the variables

-4u^2+34u-42=0

a = -4; b = 34; c = -42;
Δ = b2-4ac
Δ = 342-4·(-4)·(-42)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-22}{2*-4}=\frac{-56}{-8}
=+7
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+22}{2*-4}=\frac{-12}{-8}
=1+1/2
$