6c2+4c-13=5c2

Solution for 6c2+4c-13=5c2 equation:

6c^2+4c-13=5c^2

We move all terms to the left:

6c^2+4c-13-(5c^2)=0

determiningTheFunctionDomain
6c^2-5c^2+4c-13=0

We add all the numbers together, and all the variables

c^2+4c-13=0

a = 1; b = 4; c = -13;
Δ = b2-4ac
Δ = 42-4·1·(-13)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{17}}{2*1}=\frac{-4-2\sqrt{17}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{17}}{2*1}=\frac{-4+2\sqrt{17}}{2}
$