48-(3c+4)=(c+5)c

Solution for 48-(3c+4)=(c+5)c equation:

48-(3c+4)=(c+5)c

We move all terms to the left:

48-(3c+4)-((c+5)c)=0

We get rid of parentheses

-3c-((c+5)c)-4+48=0


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

(c+5)c

We multiply parentheses

c^2+5c

Back to the equation:

-(c^2+5c)


We add all the numbers together, and all the variables

-3c-(c^2+5c)+44=0

We get rid of parentheses

-c^2-3c-5c+44=0

We add all the numbers together, and all the variables

-1c^2-8c+44=0

a = -1; b = -8; c = +44;
Δ = b2-4ac
Δ = -82-4·(-1)·44
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{15}}{2*-1}=\frac{8-4\sqrt{15}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{15}}{2*-1}=\frac{8+4\sqrt{15}}{-2}
$