10(n^2-1)+2n-5=(2n+5)(n-1)

Solution for 10(n^2-1)+2n-5=(2n+5)(n-1) equation:

10(n^2-1)+2n-5=(2n+5)(n-1)

We move all terms to the left:

10(n^2-1)+2n-5-((2n+5)(n-1))=0

We add all the numbers together, and all the variables

2n+10(n^2-1)-((2n+5)(n-1))-5=0

We multiply parentheses

10n^2+2n-((2n+5)(n-1))-10-5=0

We multiply parentheses ..

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


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

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

We get rid of parentheses

2n^2-2n+5n-5

We add all the numbers together, and all the variables

2n^2+3n-5

Back to the equation:

-(2n^2+3n-5)


We add all the numbers together, and all the variables

10n^2+2n-(2n^2+3n-5)-15=0

We get rid of parentheses

10n^2-2n^2+2n-3n+5-15=0

We add all the numbers together, and all the variables

8n^2-1n-10=0

a = 8; b = -1; c = -10;
Δ = b2-4ac
Δ = -12-4·8·(-10)
Δ = 321
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{321}}{2*8}=\frac{1-\sqrt{321}}{16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{321}}{2*8}=\frac{1+\sqrt{321}}{16}
$