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

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

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

We multiply parentheses ..

(+2n^2-3n+6n-9)=0

We get rid of parentheses

2n^2-3n+6n-9=0

We add all the numbers together, and all the variables

2n^2+3n-9=0

a = 2; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·2·(-9)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*2}=\frac{-12}{4}
=-3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*2}=\frac{6}{4}
=1+1/2
$