35n^2=15+4n

Solution for 35n^2=15+4n equation:

35n^2=15+4n

We move all terms to the left:

35n^2-(15+4n)=0

We add all the numbers together, and all the variables

35n^2-(4n+15)=0

We get rid of parentheses

35n^2-4n-15=0

a = 35; b = -4; c = -15;
Δ = b2-4ac
Δ = -42-4·35·(-15)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-46}{2*35}=\frac{-42}{70}
=-3/5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+46}{2*35}=\frac{50}{70}
=5/7
$