1=3-2n^2+3n

Solution for 1=3-2n^2+3n equation:

1=3-2n^2+3n

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-3n-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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