3x2+14x-16=0

Solution for 3x2+14x-16=0 equation:

3x^2+14x-16=0

a = 3; b = 14; c = -16;
Δ = b2-4ac
Δ = 142-4·3·(-16)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{97}}{2*3}=\frac{-14-2\sqrt{97}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{97}}{2*3}=\frac{-14+2\sqrt{97}}{6}
$