38=3x^2+19x-2

Solution for 38=3x^2+19x-2 equation:

38=3x^2+19x-2

We move all terms to the left:

38-(3x^2+19x-2)=0

We get rid of parentheses

-3x^2-19x+2+38=0

We add all the numbers together, and all the variables

-3x^2-19x+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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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