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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4u^2+16u+7u-28)=0

We get rid of parentheses

-4u^2+16u+7u-28=0

We add all the numbers together, and all the variables

-4u^2+23u-28=0

a = -4; b = 23; c = -28;
Δ = b2-4ac
Δ = 232-4·(-4)·(-28)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-9}{2*-4}=\frac{-32}{-8}
=+4
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+9}{2*-4}=\frac{-14}{-8}
=1+3/4
$