3c2+-17c+14=0

Solution for 3c2+-17c+14=0 equation:

3c^2+-17c+14=0

We add all the numbers together, and all the variables

3c^2-17c=0

a = 3; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·3·0
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{-(-17)-17}{2*3}=\frac{0}{6}
=0
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*3}=\frac{34}{6}
=5+2/3
$