42-(3c+4)=2c(c+4)+c

Solution for 42-(3c+4)=2c(c+4)+c equation:

42-(3c+4)=2c(c+4)+c

We move all terms to the left:

42-(3c+4)-(2c(c+4)+c)=0

We get rid of parentheses

-3c-(2c(c+4)+c)-4+42=0


We calculate terms in parentheses: -(2c(c+4)+c), so:

2c(c+4)+c

We add all the numbers together, and all the variables

c+2c(c+4)

We multiply parentheses

2c^2+c+8c

We add all the numbers together, and all the variables

2c^2+9c

Back to the equation:

-(2c^2+9c)


We add all the numbers together, and all the variables

-3c-(2c^2+9c)+38=0

We get rid of parentheses

-2c^2-3c-9c+38=0

We add all the numbers together, and all the variables

-2c^2-12c+38=0

a = -2; b = -12; c = +38;
Δ = b2-4ac
Δ = -122-4·(-2)·38
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{7}}{2*-2}=\frac{12-8\sqrt{7}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{7}}{2*-2}=\frac{12+8\sqrt{7}}{-4}
$