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

Solution for (8-u)(4u-7)=0 equation:

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

We add all the numbers together, and all the variables

(-1u+8)(4u-7)=0

We multiply parentheses ..

(-4u^2+7u+32u-56)=0

We get rid of parentheses

-4u^2+7u+32u-56=0

We add all the numbers together, and all the variables

-4u^2+39u-56=0

a = -4; b = 39; c = -56;
Δ = b2-4ac
Δ = 392-4·(-4)·(-56)
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{-(39)-25}{2*-4}=\frac{-64}{-8}
=+8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+25}{2*-4}=\frac{-14}{-8}
=1+3/4
$