((n+5)(n-2))/2=72

Solution for ((n+5)(n-2))/2=72 equation:

((n+5)(n-2))/2=72

We move all terms to the left:

((n+5)(n-2))/2-(72)=0

We multiply parentheses ..

((+n^2-2n+5n-10))/2-72=0

We multiply all the terms by the denominator


((+n^2-2n+5n-10))-72*2=0


We calculate terms in parentheses: +((+n^2-2n+5n-10)), so:

(+n^2-2n+5n-10)

We get rid of parentheses

n^2-2n+5n-10

We add all the numbers together, and all the variables

n^2+3n-10

Back to the equation:

+(n^2+3n-10)


We add all the numbers together, and all the variables

(n^2+3n-10)-144=0

We get rid of parentheses

n^2+3n-10-144=0

We add all the numbers together, and all the variables

n^2+3n-154=0

a = 1; b = 3; c = -154;
Δ = b2-4ac
Δ = 32-4·1·(-154)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-25}{2*1}=\frac{-28}{2}
=-14
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+25}{2*1}=\frac{22}{2}
=11
$