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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

n+6n-(-3n(n-3)-(n-1))-15=0


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

-3n(n-3)-(n-1)

We multiply parentheses

-3n^2+9n-(n-1)

We get rid of parentheses

-3n^2+9n-n+1

We add all the numbers together, and all the variables

-3n^2+8n+1

Back to the equation:

-(-3n^2+8n+1)


We add all the numbers together, and all the variables

-(-3n^2+8n+1)+7n-15=0

We get rid of parentheses

3n^2-8n+7n-1-15=0

We add all the numbers together, and all the variables

3n^2-1n-16=0

a = 3; b = -1; c = -16;
Δ = b2-4ac
Δ = -12-4·3·(-16)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{193}}{2*3}=\frac{1-\sqrt{193}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{193}}{2*3}=\frac{1+\sqrt{193}}{6}
$