0=2u^2+u-3

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

0=2u^2+u-3

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-2u^2-u+3=0

We add all the numbers together, and all the variables

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

a = -2; b = -1; c = +3;
Δ = b2-4ac
Δ = -12-4·(-2)·3
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{-(-1)-5}{2*-2}=\frac{-4}{-4}
=1
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-2}=\frac{6}{-4}
=-1+1/2
$