38-(2c+3)=(c+5)c

Solution for 38-(2c+3)=(c+5)c equation:

38-(2c+3)=(c+5)c

We move all terms to the left:

38-(2c+3)-((c+5)c)=0

We get rid of parentheses

-2c-((c+5)c)-3+38=0


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

(c+5)c

We multiply parentheses

c^2+5c

Back to the equation:

-(c^2+5c)


We add all the numbers together, and all the variables

-2c-(c^2+5c)+35=0

We get rid of parentheses

-c^2-2c-5c+35=0

We add all the numbers together, and all the variables

-1c^2-7c+35=0

a = -1; b = -7; c = +35;
Δ = b2-4ac
Δ = -72-4·(-1)·35
Δ = 189
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:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3\sqrt{21}}{2*-1}=\frac{7-3\sqrt{21}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3\sqrt{21}}{2*-1}=\frac{7+3\sqrt{21}}{-2}
$