42-(3c+4)=2(c+7)c

Solution for 42-(3c+4)=2(c+7)c equation:

42-(3c+4)=2(c+7)c

We move all terms to the left:

42-(3c+4)-(2(c+7)c)=0

We get rid of parentheses

-3c-(2(c+7)c)-4+42=0


We calculate terms in parentheses: -(2(c+7)c), so:

2(c+7)c

We multiply parentheses

2c^2+14c

Back to the equation:

-(2c^2+14c)


We add all the numbers together, and all the variables

-3c-(2c^2+14c)+38=0

We get rid of parentheses

-2c^2-3c-14c+38=0

We add all the numbers together, and all the variables

-2c^2-17c+38=0

a = -2; b = -17; c = +38;
Δ = b2-4ac
Δ = -172-4·(-2)·38
Δ = 593
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{-(-17)-\sqrt{593}}{2*-2}=\frac{17-\sqrt{593}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{593}}{2*-2}=\frac{17+\sqrt{593}}{-4}
$