3n^2=-49+28n

Solution for 3n^2=-49+28n equation:

3n^2=-49+28n

We move all terms to the left:

3n^2-(-49+28n)=0

We add all the numbers together, and all the variables

3n^2-(28n-49)=0

We get rid of parentheses

3n^2-28n+49=0

a = 3; b = -28; c = +49;
Δ = b2-4ac
Δ = -282-4·3·49
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-14}{2*3}=\frac{14}{6}
=2+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+14}{2*3}=\frac{42}{6}
=7
$