(3n+1)(4n+1)+(n+2)(4n+1)=0

Solution for (3n+1)(4n+1)+(n+2)(4n+1)=0 equation:

(3n+1)(4n+1)+(n+2)(4n+1)=0

We multiply parentheses ..

(+12n^2+3n+4n+1)+(n+2)(4n+1)=0

We get rid of parentheses

12n^2+3n+4n+(n+2)(4n+1)+1=0

We multiply parentheses ..

12n^2+(+4n^2+n+8n+2)+3n+4n+1=0

We add all the numbers together, and all the variables

12n^2+(+4n^2+n+8n+2)+7n+1=0

We get rid of parentheses

12n^2+4n^2+n+8n+7n+2+1=0

We add all the numbers together, and all the variables

16n^2+16n+3=0

a = 16; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·16·3
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*16}=\frac{-24}{32}
=-3/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*16}=\frac{-8}{32}
=-1/4
$