(7+u)(4u-5)=0

Solution for (7+u)(4u-5)=0 equation:

(7+u)(4u-5)=0

We add all the numbers together, and all the variables

(u+7)(4u-5)=0

We multiply parentheses ..

(+4u^2-5u+28u-35)=0

We get rid of parentheses

4u^2-5u+28u-35=0

We add all the numbers together, and all the variables

4u^2+23u-35=0

a = 4; b = 23; c = -35;
Δ = b2-4ac
Δ = 232-4·4·(-35)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-33}{2*4}=\frac{-56}{8}
=-7
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+33}{2*4}=\frac{10}{8}
=1+1/4
$