c+7=5c(3+4c)

Solution for c+7=5c(3+4c) equation:

c+7=5c(3+4c)

We move all terms to the left:

c+7-(5c(3+4c))=0

We add all the numbers together, and all the variables

c-(5c(4c+3))+7=0


We calculate terms in parentheses: -(5c(4c+3)), so:

5c(4c+3)

We multiply parentheses

20c^2+15c

Back to the equation:

-(20c^2+15c)


We get rid of parentheses

-20c^2+c-15c+7=0

We add all the numbers together, and all the variables

-20c^2-14c+7=0

a = -20; b = -14; c = +7;
Δ = b2-4ac
Δ = -142-4·(-20)·7
Δ = 756
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:

$\sqrt{\Delta}=\sqrt{756}=\sqrt{36*21}=\sqrt{36}*\sqrt{21}=6\sqrt{21}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{21}}{2*-20}=\frac{14-6\sqrt{21}}{-40}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{21}}{2*-20}=\frac{14+6\sqrt{21}}{-40}
$