(2n+10)(2n-4)=228

Solution for (2n+10)(2n-4)=228 equation:

(2n+10)(2n-4)=228

We move all terms to the left:

(2n+10)(2n-4)-(228)=0

We multiply parentheses ..

(+4n^2-8n+20n-40)-228=0

We get rid of parentheses

4n^2-8n+20n-40-228=0

We add all the numbers together, and all the variables

4n^2+12n-268=0

a = 4; b = 12; c = -268;
Δ = b2-4ac
Δ = 122-4·4·(-268)
Δ = 4432
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:

$\sqrt{\Delta}=\sqrt{4432}=\sqrt{16*277}=\sqrt{16}*\sqrt{277}=4\sqrt{277}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{277}}{2*4}=\frac{-12-4\sqrt{277}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{277}}{2*4}=\frac{-12+4\sqrt{277}}{8}
$