3c^2-18c+26=0

Solution for 3c^2-18c+26=0 equation:

3c^2-18c+26=0

a = 3; b = -18; c = +26;
Δ = b2-4ac
Δ = -182-4·3·26
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{3}}{2*3}=\frac{18-2\sqrt{3}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{3}}{2*3}=\frac{18+2\sqrt{3}}{6}
$