34-(2c+4)=2(c+5)c

Solution for 34-(2c+4)=2(c+5)c equation:

34-(2c+4)=2(c+5)c

We move all terms to the left:

34-(2c+4)-(2(c+5)c)=0

We get rid of parentheses

-2c-(2(c+5)c)-4+34=0


We calculate terms in parentheses: -(2(c+5)c), so:

2(c+5)c

We multiply parentheses

2c^2+10c

Back to the equation:

-(2c^2+10c)


We add all the numbers together, and all the variables

-2c-(2c^2+10c)+30=0

We get rid of parentheses

-2c^2-2c-10c+30=0

We add all the numbers together, and all the variables

-2c^2-12c+30=0

a = -2; b = -12; c = +30;
Δ = b2-4ac
Δ = -122-4·(-2)·30
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{6}}{2*-2}=\frac{12-8\sqrt{6}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{6}}{2*-2}=\frac{12+8\sqrt{6}}{-4}
$