3c^2+2c-16=0

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

3c^2+2c-16=0

a = 3; b = 2; c = -16;
Δ = b2-4ac
Δ = 22-4·3·(-16)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*3}=\frac{-16}{6}
=-2+2/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*3}=\frac{12}{6}
=2
$