(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

(u+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+9u-35=0

a = 2; b = 9; c = -35;
Δ = b2-4ac
Δ = 92-4·2·(-35)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-19}{2*2}=\frac{-28}{4}
=-7
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+19}{2*2}=\frac{10}{4}
=2+1/2
$