2(u-3)(u+1)=0

Solution for 2(u-3)(u+1)=0 equation:

2(u-3)(u+1)=0

We multiply parentheses ..

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

We multiply parentheses

2u^2+2u-6u-6=0

We add all the numbers together, and all the variables

2u^2-4u-6=0

a = 2; b = -4; c = -6;
Δ = b2-4ac
Δ = -42-4·2·(-6)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*2}=\frac{-4}{4}
=-1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*2}=\frac{12}{4}
=3
$