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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-5n-12=0

a = 2; b = -5; c = -12;
Δ = b2-4ac
Δ = -52-4·2·(-12)
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*2}=\frac{-6}{4}
=-1+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*2}=\frac{16}{4}
=4
$