n(2n-3)=44

Solution for n(2n-3)=44 equation:

n(2n-3)=44

We move all terms to the left:

n(2n-3)-(44)=0

We multiply parentheses

2n^2-3n-44=0

a = 2; b = -3; c = -44;
Δ = b2-4ac
Δ = -32-4·2·(-44)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-19}{2*2}=\frac{-16}{4}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+19}{2*2}=\frac{22}{4}
=5+1/2
$