(3n2-7n)+(7n2+n=1)

Solution for (3n2-7n)+(7n2+n=1) equation:

(3n^2-7n)+(7n^2+n=1)

We move all terms to the left:

(3n^2-7n)+(7n^2+n-(1))=0

We get rid of parentheses

3n^2+7n^2-7n+n-1=0

We add all the numbers together, and all the variables

10n^2-6n-1=0

a = 10; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·10·(-1)
Δ = 76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{19}}{2*10}=\frac{6-2\sqrt{19}}{20}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{19}}{2*10}=\frac{6+2\sqrt{19}}{20}
$