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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2+3n-7=0

a = 2; b = 3; c = -7;
Δ = b2-4ac
Δ = 32-4·2·(-7)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{65}}{2*2}=\frac{-3-\sqrt{65}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{65}}{2*2}=\frac{-3+\sqrt{65}}{4}
$