(7-u)(2u-5)=0

Solution for (7-u)(2u-5)=0 equation:

(7-u)(2u-5)=0

We add all the numbers together, and all the variables

(-1u+7)(2u-5)=0

We multiply parentheses ..

(-2u^2+5u+14u-35)=0

We get rid of parentheses

-2u^2+5u+14u-35=0

We add all the numbers together, and all the variables

-2u^2+19u-35=0

a = -2; b = 19; c = -35;
Δ = b2-4ac
Δ = 192-4·(-2)·(-35)
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{-(19)-9}{2*-2}=\frac{-28}{-4}
=+7
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+9}{2*-2}=\frac{-10}{-4}
=2+1/2
$