(2u-3)(6+u)=0

Solution for (2u-3)(6+u)=0 equation:

(2u-3)(6+u)=0

We add all the numbers together, and all the variables

(2u-3)(u+6)=0

We multiply parentheses ..

(+2u^2+12u-3u-18)=0

We get rid of parentheses

2u^2+12u-3u-18=0

We add all the numbers together, and all the variables

2u^2+9u-18=0

a = 2; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·2·(-18)
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{-(9)-15}{2*2}=\frac{-24}{4}
=-6
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*2}=\frac{6}{4}
=1+1/2
$