33-(2c+4)=3c(c+7)

Solution for 33-(2c+4)=3c(c+7) equation:

33-(2c+4)=3c(c+7)

We move all terms to the left:

33-(2c+4)-(3c(c+7))=0

We get rid of parentheses

-2c-(3c(c+7))-4+33=0


We calculate terms in parentheses: -(3c(c+7)), so:

3c(c+7)

We multiply parentheses

3c^2+21c

Back to the equation:

-(3c^2+21c)


We add all the numbers together, and all the variables

-2c-(3c^2+21c)+29=0

We get rid of parentheses

-3c^2-2c-21c+29=0

We add all the numbers together, and all the variables

-3c^2-23c+29=0

a = -3; b = -23; c = +29;
Δ = b2-4ac
Δ = -232-4·(-3)·29
Δ = 877
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{-(-23)-\sqrt{877}}{2*-3}=\frac{23-\sqrt{877}}{-6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{877}}{2*-3}=\frac{23+\sqrt{877}}{-6}
$