-3u+2u^2=3

Solution for -3u+2u^2=3 equation:

-3u+2u^2=3

We move all terms to the left:

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

a = 2; b = -3; c = -3;
Δ = b2-4ac
Δ = -32-4·2·(-3)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*2}=\frac{3-\sqrt{33}}{4}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*2}=\frac{3+\sqrt{33}}{4}
$