-3x^2-16x-4=0

Solution for -3x^2-16x-4=0 equation:

-3x^2-16x-4=0

a = -3; b = -16; c = -4;
Δ = b2-4ac
Δ = -162-4·(-3)·(-4)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{13}}{2*-3}=\frac{16-4\sqrt{13}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{13}}{2*-3}=\frac{16+4\sqrt{13}}{-6}
$