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

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

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

We multiply parentheses ..

(+6n^2-8n+9n-12)=0

We get rid of parentheses

6n^2-8n+9n-12=0

We add all the numbers together, and all the variables

6n^2+n-12=0

a = 6; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·6·(-12)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*6}=\frac{-18}{12}
=-1+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*6}=\frac{16}{12}
=1+1/3
$