44-(3c+4)=4(c+6)c

Solution for 44-(3c+4)=4(c+6)c equation:

44-(3c+4)=4(c+6)c

We move all terms to the left:

44-(3c+4)-(4(c+6)c)=0

We get rid of parentheses

-3c-(4(c+6)c)-4+44=0


We calculate terms in parentheses: -(4(c+6)c), so:

4(c+6)c

We multiply parentheses

4c^2+24c

Back to the equation:

-(4c^2+24c)


We add all the numbers together, and all the variables

-3c-(4c^2+24c)+40=0

We get rid of parentheses

-4c^2-3c-24c+40=0

We add all the numbers together, and all the variables

-4c^2-27c+40=0

a = -4; b = -27; c = +40;
Δ = b2-4ac
Δ = -272-4·(-4)·40
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-37}{2*-4}=\frac{-10}{-8}
=1+1/4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+37}{2*-4}=\frac{64}{-8}
=-8
$