3n2-19n=40

Solution for 3n2-19n=40 equation:

3n^2-19n=40

We move all terms to the left:

3n^2-19n-(40)=0

a = 3; b = -19; c = -40;
Δ = b2-4ac
Δ = -192-4·3·(-40)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-29}{2*3}=\frac{-10}{6}
=-1+2/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+29}{2*3}=\frac{48}{6}
=8
$