(8+u)(2u+1)=0

Solution for (8+u)(2u+1)=0 equation:

(8+u)(2u+1)=0

We add all the numbers together, and all the variables

(u+8)(2u+1)=0

We multiply parentheses ..

(+2u^2+u+16u+8)=0

We get rid of parentheses

2u^2+u+16u+8=0

We add all the numbers together, and all the variables

2u^2+17u+8=0

a = 2; b = 17; c = +8;
Δ = b2-4ac
Δ = 172-4·2·8
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-15}{2*2}=\frac{-32}{4}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+15}{2*2}=\frac{-2}{4}
=-1/2
$