(3-2n)(n-4)=0

Solution for (3-2n)(n-4)=0 equation:

(3-2n)(n-4)=0

We add all the numbers together, and all the variables

(-2n+3)(n-4)=0

We multiply parentheses ..

(-2n^2+8n+3n-12)=0

We get rid of parentheses

-2n^2+8n+3n-12=0

We add all the numbers together, and all the variables

-2n^2+11n-12=0

a = -2; b = 11; c = -12;
Δ = b2-4ac
Δ = 112-4·(-2)·(-12)
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{-(11)-5}{2*-2}=\frac{-16}{-4}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*-2}=\frac{-6}{-4}
=1+1/2
$