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

Solution for -3x^2+19x+40=0 equation:

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