-3x^2-6x+16=0

Solution for -3x^2-6x+16=0 equation:

-3x^2-6x+16=0

a = -3; b = -6; c = +16;
Δ = b2-4ac
Δ = -62-4·(-3)·16
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{57}}{2*-3}=\frac{6-2\sqrt{57}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{57}}{2*-3}=\frac{6+2\sqrt{57}}{-6}
$