2n-2(4n+5)=-6n(n-3)-8

Solution for 2n-2(4n+5)=-6n(n-3)-8 equation:

2n-2(4n+5)=-6n(n-3)-8

We move all terms to the left:

2n-2(4n+5)-(-6n(n-3)-8)=0

We multiply parentheses

2n-8n-(-6n(n-3)-8)-10=0


We calculate terms in parentheses: -(-6n(n-3)-8), so:

-6n(n-3)-8

We multiply parentheses

-6n^2+18n-8

Back to the equation:

-(-6n^2+18n-8)


We add all the numbers together, and all the variables

-(-6n^2+18n-8)-6n-10=0

We get rid of parentheses

6n^2-18n-6n+8-10=0

We add all the numbers together, and all the variables

6n^2-24n-2=0

a = 6; b = -24; c = -2;
Δ = b2-4ac
Δ = -242-4·6·(-2)
Δ = 624
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:

$\sqrt{\Delta}=\sqrt{624}=\sqrt{16*39}=\sqrt{16}*\sqrt{39}=4\sqrt{39}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{39}}{2*6}=\frac{24-4\sqrt{39}}{12}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{39}}{2*6}=\frac{24+4\sqrt{39}}{12}
$