40-(3c+4)=2(c+6)c

Solution for 40-(3c+4)=2(c+6)c equation:

40-(3c+4)=2(c+6)c

We move all terms to the left:

40-(3c+4)-(2(c+6)c)=0

We get rid of parentheses

-3c-(2(c+6)c)-4+40=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)+36=0

We get rid of parentheses

-2c^2-3c-12c+36=0

We add all the numbers together, and all the variables

-2c^2-15c+36=0

a = -2; b = -15; c = +36;
Δ = b2-4ac
Δ = -152-4·(-2)·36
Δ = 513
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:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{57}}{2*-2}=\frac{15-3\sqrt{57}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{57}}{2*-2}=\frac{15+3\sqrt{57}}{-4}
$