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

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

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

a = -3; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·(-3)·3
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{73}}{2*-3}=\frac{-16-2\sqrt{73}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{73}}{2*-3}=\frac{-16+2\sqrt{73}}{-6}
$