4n(3+2n)=21

Solution for 4n(3+2n)=21 equation:

4n(3+2n)=21

We move all terms to the left:

4n(3+2n)-(21)=0

We add all the numbers together, and all the variables

4n(2n+3)-21=0

We multiply parentheses

8n^2+12n-21=0

a = 8; b = 12; c = -21;
Δ = b2-4ac
Δ = 122-4·8·(-21)
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{51}}{2*8}=\frac{-12-4\sqrt{51}}{16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{51}}{2*8}=\frac{-12+4\sqrt{51}}{16}
$