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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4n^2-8n+3n+6)=0

We get rid of parentheses

-4n^2-8n+3n+6=0

We add all the numbers together, and all the variables

-4n^2-5n+6=0

a = -4; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·(-4)·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-4}=\frac{-6}{-8}
=3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-4}=\frac{16}{-8}
=-2
$