(4n-3)(2n-1)=63

Solution for (4n-3)(2n-1)=63 equation:

(4n-3)(2n-1)=63

We move all terms to the left:

(4n-3)(2n-1)-(63)=0

We multiply parentheses ..

(+8n^2-4n-6n+3)-63=0

We get rid of parentheses

8n^2-4n-6n+3-63=0

We add all the numbers together, and all the variables

8n^2-10n-60=0

a = 8; b = -10; c = -60;
Δ = b2-4ac
Δ = -102-4·8·(-60)
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{505}}{2*8}=\frac{10-2\sqrt{505}}{16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{505}}{2*8}=\frac{10+2\sqrt{505}}{16}
$