16c+2=c^2+c-2c

Solution for 16c+2=c^2+c-2c equation:

16c+2=c^2+c-2c

We move all terms to the left:

16c+2-(c^2+c-2c)=0

We get rid of parentheses

-c^2+16c-c+2c+2=0

We add all the numbers together, and all the variables

-1c^2+17c+2=0

a = -1; b = 17; c = +2;
Δ = b2-4ac
Δ = 172-4·(-1)·2
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-3\sqrt{33}}{2*-1}=\frac{-17-3\sqrt{33}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+3\sqrt{33}}{2*-1}=\frac{-17+3\sqrt{33}}{-2}
$