4n2=5-19n

Solution for 4n2=5-19n equation:

4n^2=5-19n

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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