-3c^2+16c-5=0

Solution for -3c^2+16c-5=0 equation:

-3c^2+16c-5=0

a = -3; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·(-3)·(-5)
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{-(16)-14}{2*-3}=\frac{-30}{-6}
=+5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*-3}=\frac{-2}{-6}
=1/3
$