3n2=21n+24

Solution for 3n2=21n+24 equation:

3n^2=21n+24

We move all terms to the left:

3n^2-(21n+24)=0

We get rid of parentheses

3n^2-21n-24=0

a = 3; b = -21; c = -24;
Δ = b2-4ac
Δ = -212-4·3·(-24)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-27}{2*3}=\frac{-6}{6}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+27}{2*3}=\frac{48}{6}
=8
$