4n(1+3n)-7n(n-4)=

Solution for 4n(1+3n)-7n(n-4)= equation:

4n(1+3n)-7n(n-4)=

We move all terms to the left:

4n(1+3n)-7n(n-4)-()=0

We add all the numbers together, and all the variables

4n(3n+1)-7n(n-4)-()=0

We add all the numbers together, and all the variables

4n(3n+1)-7n(n-4)=0

We multiply parentheses

12n^2-7n^2+4n+28n=0

We add all the numbers together, and all the variables

5n^2+32n=0

a = 5; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·5·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*5}=\frac{-64}{10}
=-6+2/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*5}=\frac{0}{10}
=0
$