3x^2=19x+40

Solution for 3x^2=19x+40 equation:

3x^2=19x+40

We move all terms to the left:

3x^2-(19x+40)=0

We get rid of parentheses

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
$