-3x^2+24x+3840=0

Solution for -3x^2+24x+3840=0 equation:

-3x^2+24x+3840=0

a = -3; b = 24; c = +3840;
Δ = b2-4ac
Δ = 242-4·(-3)·3840
Δ = 46656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{46656}=216$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-216}{2*-3}=\frac{-240}{-6}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+216}{2*-3}=\frac{192}{-6}
=-32
$