3n(n+4)=21

Solution for 3n(n+4)=21 equation:

3n(n+4)=21

We move all terms to the left:

3n(n+4)-(21)=0

We multiply parentheses

3n^2+12n-21=0

a = 3; b = 12; c = -21;
Δ = b2-4ac
Δ = 122-4·3·(-21)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{11}}{2*3}=\frac{-12-6\sqrt{11}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{11}}{2*3}=\frac{-12+6\sqrt{11}}{6}
$