c2=3c+14

Solution for c2=3c+14 equation:

c2=3c+14

We move all terms to the left:

c2-(3c+14)=0

We add all the numbers together, and all the variables

c^2-(3c+14)=0

We get rid of parentheses

c^2-3c-14=0

a = 1; b = -3; c = -14;
Δ = b2-4ac
Δ = -32-4·1·(-14)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{65}}{2*1}=\frac{3-\sqrt{65}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{65}}{2*1}=\frac{3+\sqrt{65}}{2}
$