5c+4c(c-1)=2+5(c+2)

Solution for 5c+4c(c-1)=2+5(c+2) equation:

5c+4c(c-1)=2+5(c+2)

We move all terms to the left:

5c+4c(c-1)-(2+5(c+2))=0

We multiply parentheses

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


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

2+5(c+2)

determiningTheFunctionDomain
5(c+2)+2

We multiply parentheses

5c+10+2

We add all the numbers together, and all the variables

5c+12

Back to the equation:

-(5c+12)


We add all the numbers together, and all the variables

4c^2+c-(5c+12)=0

We get rid of parentheses

4c^2+c-5c-12=0

We add all the numbers together, and all the variables

4c^2-4c-12=0

a = 4; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·4·(-12)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{13}}{2*4}=\frac{4-4\sqrt{13}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{13}}{2*4}=\frac{4+4\sqrt{13}}{8}
$