(2u-3)(8+u)=0

Solution for (2u-3)(8+u)=0 equation:

(2u-3)(8+u)=0

We add all the numbers together, and all the variables

(2u-3)(u+8)=0

We multiply parentheses ..

(+2u^2+16u-3u-24)=0

We get rid of parentheses

2u^2+16u-3u-24=0

We add all the numbers together, and all the variables

2u^2+13u-24=0

a = 2; b = 13; c = -24;
Δ = b2-4ac
Δ = 132-4·2·(-24)
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{-(13)-19}{2*2}=\frac{-32}{4}
=-8
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*2}=\frac{6}{4}
=1+1/2
$