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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-2n^2+n+6n-3)=0

We get rid of parentheses

-2n^2+n+6n-3=0

We add all the numbers together, and all the variables

-2n^2+7n-3=0

a = -2; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·(-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*-2}=\frac{-12}{-4}
=+3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*-2}=\frac{-2}{-4}
=1/2
$