-3c^2+13c+10=0

Solution for -3c^2+13c+10=0 equation:

-3c^2+13c+10=0

a = -3; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·(-3)·10
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*-3}=\frac{-30}{-6}
=+5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-3}=\frac{4}{-6}
=-2/3
$