3x2+6x-16=0

Solution for 3x2+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}
$