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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


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

4(c+5)c

We multiply parentheses

4c^2+20c

Back to the equation:

-(4c^2+20c)


We add all the numbers together, and all the variables

-3c-(4c^2+20c)+44=0

We get rid of parentheses

-4c^2-3c-20c+44=0

We add all the numbers together, and all the variables

-4c^2-23c+44=0

a = -4; b = -23; c = +44;
Δ = b2-4ac
Δ = -232-4·(-4)·44
Δ = 1233
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:

$\sqrt{\Delta}=\sqrt{1233}=\sqrt{9*137}=\sqrt{9}*\sqrt{137}=3\sqrt{137}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-3\sqrt{137}}{2*-4}=\frac{23-3\sqrt{137}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+3\sqrt{137}}{2*-4}=\frac{23+3\sqrt{137}}{-8}
$