6(2n-3)+n(2-6)=3(n-9n2)+2(n+7)

Solution for 6(2n-3)+n(2-6)=3(n-9n2)+2(n+7) equation:

6(2n-3)+n(2-6)=3(n-9n^2)+2(n+7)

We move all terms to the left:

6(2n-3)+n(2-6)-(3(n-9n^2)+2(n+7))=0

We add all the numbers together, and all the variables

-(3(n-9n^2)+2(n+7))+6(2n-3)+n(-4)=0

We multiply parentheses

-(3(n-9n^2)+2(n+7))+12n-4n-18=0


We calculate terms in parentheses: -(3(n-9n^2)+2(n+7)), so:

3(n-9n^2)+2(n+7)

We multiply parentheses

-27n^2+3n+2n+14

We add all the numbers together, and all the variables

-27n^2+5n+14

Back to the equation:

-(-27n^2+5n+14)


We add all the numbers together, and all the variables

-(-27n^2+5n+14)+8n-18=0

We get rid of parentheses

27n^2-5n+8n-14-18=0

We add all the numbers together, and all the variables

27n^2+3n-32=0

a = 27; b = 3; c = -32;
Δ = b2-4ac
Δ = 32-4·27·(-32)
Δ = 3465
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:

$\sqrt{\Delta}=\sqrt{3465}=\sqrt{9*385}=\sqrt{9}*\sqrt{385}=3\sqrt{385}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{385}}{2*27}=\frac{-3-3\sqrt{385}}{54}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{385}}{2*27}=\frac{-3+3\sqrt{385}}{54}
$