n(2n-5)=-1

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

n(2n-5)=-1

We move all terms to the left:

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

We add all the numbers together, and all the variables

n(2n-5)+1=0

We multiply parentheses

2n^2-5n+1=0

a = 2; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·2·1
Δ = 17
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{-(-5)-\sqrt{17}}{2*2}=\frac{5-\sqrt{17}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{17}}{2*2}=\frac{5+\sqrt{17}}{4}
$