(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

(u+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+25u-56=0

a = 4; b = 25; c = -56;
Δ = b2-4ac
Δ = 252-4·4·(-56)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-39}{2*4}=\frac{-64}{8}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+39}{2*4}=\frac{14}{8}
=1+3/4
$