-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 = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{19}}{2*-3}=\frac{-16-4\sqrt{19}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{19}}{2*-3}=\frac{-16+4\sqrt{19}}{-6}
$