-4n(n-2)=6(n+3)-11n^2

Solution for -4n(n-2)=6(n+3)-11n^2 equation:

-4n(n-2)=6(n+3)-11n^2

We move all terms to the left:

-4n(n-2)-(6(n+3)-11n^2)=0

We multiply parentheses

-(6(n+3)-11n^2)-4n^2+8n=0


We calculate terms in parentheses: -(6(n+3)-11n^2), so:

6(n+3)-11n^2

determiningTheFunctionDomain
-11n^2+6(n+3)

We multiply parentheses

-11n^2+6n+18

Back to the equation:

-(-11n^2+6n+18)


We add all the numbers together, and all the variables

-4n^2-(-11n^2+6n+18)+8n=0

We get rid of parentheses

-4n^2+11n^2-6n+8n-18=0

We add all the numbers together, and all the variables

7n^2+2n-18=0

a = 7; b = 2; c = -18;
Δ = b2-4ac
Δ = 22-4·7·(-18)
Δ = 508
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:

$\sqrt{\Delta}=\sqrt{508}=\sqrt{4*127}=\sqrt{4}*\sqrt{127}=2\sqrt{127}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{127}}{2*7}=\frac{-2-2\sqrt{127}}{14}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{127}}{2*7}=\frac{-2+2\sqrt{127}}{14}
$