n+5n+7=2(2n+1)3n-2

Solution for n+5n+7=2(2n+1)3n-2 equation:

n+5n+7=2(2n+1)3n-2

We move all terms to the left:

n+5n+7-(2(2n+1)3n-2)=0

We add all the numbers together, and all the variables

6n-(2(2n+1)3n-2)+7=0


We calculate terms in parentheses: -(2(2n+1)3n-2), so:

2(2n+1)3n-2

We multiply parentheses

12n^2+6n-2

Back to the equation:

-(12n^2+6n-2)


We get rid of parentheses

-12n^2+6n-6n+2+7=0

We add all the numbers together, and all the variables

-12n^2+9=0

a = -12; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-12)·9
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*-12}=\frac{0-12\sqrt{3}}{-24}
=-\frac{12\sqrt{3}}{-24}
=-\frac{\sqrt{3}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*-12}=\frac{0+12\sqrt{3}}{-24}
=\frac{12\sqrt{3}}{-24}
=\frac{\sqrt{3}}{-2}
$