(3n+4)(n+4)=0

Solution for (3n+4)(n+4)=0 equation:

(3n+4)(n+4)=0

We multiply parentheses ..

(+3n^2+12n+4n+16)=0

We get rid of parentheses

3n^2+12n+4n+16=0

We add all the numbers together, and all the variables

3n^2+16n+16=0

a = 3; b = 16; c = +16;
Δ = b2-4ac
Δ = 162-4·3·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*3}=\frac{-24}{6}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*3}=\frac{-8}{6}
=-1+1/3
$