35-19n=-2n^2

Solution for 35-19n=-2n^2 equation:

35-19n=-2n^2

We move all terms to the left:

35-19n-(-2n^2)=0

We get rid of parentheses

2n^2-19n+35=0

a = 2; b = -19; c = +35;
Δ = b2-4ac
Δ = -192-4·2·35
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-9}{2*2}=\frac{10}{4}
=2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+9}{2*2}=\frac{28}{4}
=7
$