10n^2-3n^2=14n+11n

Solution for 10n^2-3n^2=14n+11n equation:

10n^2-3n^2=14n+11n

We move all terms to the left:

10n^2-3n^2-(14n+11n)=0

We add all the numbers together, and all the variables

10n^2-3n^2-(+25n)=0

We add all the numbers together, and all the variables

7n^2-(+25n)=0

We get rid of parentheses

7n^2-25n=0

a = 7; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·7·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*7}=\frac{0}{14}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*7}=\frac{50}{14}
=3+4/7
$