(n+4)(n-3)/2=9

Solution for (n+4)(n-3)/2=9 equation:

(n+4)(n-3)/2=9

We move all terms to the left:

(n+4)(n-3)/2-(9)=0

We multiply parentheses ..

(+n^2-3n+4n-12)/2-9=0

We multiply all the terms by the denominator


(+n^2-3n+4n-12)-9*2=0

We add all the numbers together, and all the variables

(+n^2-3n+4n-12)-18=0

We get rid of parentheses

n^2-3n+4n-12-18=0

We add all the numbers together, and all the variables

n^2+n-30=0

a = 1; b = 1; c = -30;
Δ = b2-4ac
Δ = 12-4·1·(-30)
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{-(1)-11}{2*1}=\frac{-12}{2}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*1}=\frac{10}{2}
=5
$