-1+2+5+8+(3n-4)=(n/2)(3n-5)

Solution for -1+2+5+8+(3n-4)=(n/2)(3n-5) equation:

-1+2+5+8+(3n-4)=(n/2)(3n-5)

We move all terms to the left:

-1+2+5+8+(3n-4)-((n/2)(3n-5))=0


Domain of the equation: 2)(3n-5))!=0

n∈R


We add all the numbers together, and all the variables

(3n-4)-((+n/2)(3n-5))-1+2+5+8=0

We add all the numbers together, and all the variables

(3n-4)-((+n/2)(3n-5))+14=0

We get rid of parentheses

3n-((+n/2)(3n-5))-4+14=0

We multiply parentheses ..

-((+3n^2-5n))+3n-4+14=0


We calculate terms in parentheses: -((+3n^2-5n)), so:

(+3n^2-5n)

We get rid of parentheses

3n^2-5n

Back to the equation:

-(3n^2-5n)


We add all the numbers together, and all the variables

3n-(3n^2-5n)+10=0

We get rid of parentheses

-3n^2+3n+5n+10=0

We add all the numbers together, and all the variables

-3n^2+8n+10=0

a = -3; b = 8; c = +10;
Δ = b2-4ac
Δ = 82-4·(-3)·10
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{46}}{2*-3}=\frac{-8-2\sqrt{46}}{-6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{46}}{2*-3}=\frac{-8+2\sqrt{46}}{-6}
$