7=c^2+6c

Solution for 7=c^2+6c equation:

7=c^2+6c

We move all terms to the left:

7-(c^2+6c)=0

We get rid of parentheses

-c^2-6c+7=0

We add all the numbers together, and all the variables

-1c^2-6c+7=0

a = -1; b = -6; c = +7;
Δ = b2-4ac
Δ = -62-4·(-1)·7
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-8}{2*-1}=\frac{-2}{-2}
=1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+8}{2*-1}=\frac{14}{-2}
=-7
$