2n^2+14=11n

Solution for 2n^2+14=11n equation:

2n^2+14=11n

We move all terms to the left:

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

a = 2; b = -11; c = +14;
Δ = b2-4ac
Δ = -112-4·2·14
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-3}{2*2}=\frac{8}{4}
=2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+3}{2*2}=\frac{14}{4}
=3+1/2
$