(2n)2=10n+48

Solution for (2n)2=10n+48 equation:

(2n)2=10n+48

We move all terms to the left:

(2n)2-(10n+48)=0

We add all the numbers together, and all the variables

2n^2-(10n+48)=0

We get rid of parentheses

2n^2-10n-48=0

a = 2; b = -10; c = -48;
Δ = b2-4ac
Δ = -102-4·2·(-48)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*2}=\frac{-12}{4}
=-3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*2}=\frac{32}{4}
=8
$