24=3(n-5)n=

Solution for 24=3(n-5)n= equation:

24=3(n-5)n=

We move all terms to the left:

24-(3(n-5)n)=0


We calculate terms in parentheses: -(3(n-5)n), so:

3(n-5)n

We multiply parentheses

3n^2-15n

Back to the equation:

-(3n^2-15n)


We get rid of parentheses

-3n^2+15n+24=0

a = -3; b = 15; c = +24;
Δ = b2-4ac
Δ = 152-4·(-3)·24
Δ = 513
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{57}}{2*-3}=\frac{-15-3\sqrt{57}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{57}}{2*-3}=\frac{-15+3\sqrt{57}}{-6}
$