16n2+32=4(n+8)

Solution for 16n2+32=4(n+8) equation:

16n^2+32=4(n+8)

We move all terms to the left:

16n^2+32-(4(n+8))=0


We calculate terms in parentheses: -(4(n+8)), so:

4(n+8)

We multiply parentheses

4n+32

Back to the equation:

-(4n+32)


We get rid of parentheses

16n^2-4n-32+32=0

We add all the numbers together, and all the variables

16n^2-4n=0

a = 16; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·16·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*16}=\frac{0}{32}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*16}=\frac{8}{32}
=1/4
$