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

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

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

We multiply parentheses

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

a = -2; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-2)·2
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*-2}=\frac{-8}{-4}
=+2
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*-2}=\frac{2}{-4}
=-1/2
$