28-(3c+4)=2(c+6)c

Solution for 28-(3c+4)=2(c+6)c equation:

28-(3c+4)=2(c+6)c

We move all terms to the left:

28-(3c+4)-(2(c+6)c)=0

We get rid of parentheses

-3c-(2(c+6)c)-4+28=0


We calculate terms in parentheses: -(2(c+6)c), so:

2(c+6)c

We multiply parentheses

2c^2+12c

Back to the equation:

-(2c^2+12c)


We add all the numbers together, and all the variables

-3c-(2c^2+12c)+24=0

We get rid of parentheses

-2c^2-3c-12c+24=0

We add all the numbers together, and all the variables

-2c^2-15c+24=0

a = -2; b = -15; c = +24;
Δ = b2-4ac
Δ = -152-4·(-2)·24
Δ = 417
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{417}}{2*-2}=\frac{15-\sqrt{417}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{417}}{2*-2}=\frac{15+\sqrt{417}}{-4}
$