4n^2-5=19n

Solution for 4n^2-5=19n equation:

4n^2-5=19n

We move all terms to the left:

4n^2-5-(19n)=0

a = 4; b = -19; c = -5;
Δ = b2-4ac
Δ = -192-4·4·(-5)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*4}=\frac{-2}{8}
=-1/4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*4}=\frac{40}{8}
=5
$