4n^2=160+12n

Solution for 4n^2=160+12n equation:

4n^2=160+12n

We move all terms to the left:

4n^2-(160+12n)=0

We add all the numbers together, and all the variables

4n^2-(12n+160)=0

We get rid of parentheses

4n^2-12n-160=0

a = 4; b = -12; c = -160;
Δ = b2-4ac
Δ = -122-4·4·(-160)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-52}{2*4}=\frac{-40}{8}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+52}{2*4}=\frac{64}{8}
=8
$