19-(2c+3)=2c(c+3)+c

Solution for 19-(2c+3)=2c(c+3)+c equation:

19-(2c+3)=2c(c+3)+c

We move all terms to the left:

19-(2c+3)-(2c(c+3)+c)=0

We get rid of parentheses

-2c-(2c(c+3)+c)-3+19=0


We calculate terms in parentheses: -(2c(c+3)+c), so:

2c(c+3)+c

We add all the numbers together, and all the variables

c+2c(c+3)

We multiply parentheses

2c^2+c+6c

We add all the numbers together, and all the variables

2c^2+7c

Back to the equation:

-(2c^2+7c)


We add all the numbers together, and all the variables

-2c-(2c^2+7c)+16=0

We get rid of parentheses

-2c^2-2c-7c+16=0

We add all the numbers together, and all the variables

-2c^2-9c+16=0

a = -2; b = -9; c = +16;
Δ = b2-4ac
Δ = -92-4·(-2)·16
Δ = 209
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{-(-9)-\sqrt{209}}{2*-2}=\frac{9-\sqrt{209}}{-4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{209}}{2*-2}=\frac{9+\sqrt{209}}{-4}
$