-3x^2-24x+16=0

Solution for -3x^2-24x+16=0 equation:

-3x^2-24x+16=0

a = -3; b = -24; c = +16;
Δ = b2-4ac
Δ = -242-4·(-3)·16
Δ = 768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16\sqrt{3}}{2*-3}=\frac{24-16\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16\sqrt{3}}{2*-3}=\frac{24+16\sqrt{3}}{-6}
$