2(3+n)n=4

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

2(3+n)n=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

2(n+3)n-4=0

We multiply parentheses

2n^2+6n-4=0

a = 2; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·2·(-4)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{17}}{2*2}=\frac{-6-2\sqrt{17}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{17}}{2*2}=\frac{-6+2\sqrt{17}}{4}
$