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

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

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

We multiply parentheses

6u^2+2u-4=0

a = 6; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·6·(-4)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*6}=\frac{-12}{12}
=-1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*6}=\frac{8}{12}
=2/3
$