3c(c-2)=5(c+1)

Solution for 3c(c-2)=5(c+1) equation:

3c(c-2)=5(c+1)

We move all terms to the left:

3c(c-2)-(5(c+1))=0

We multiply parentheses

3c^2-6c-(5(c+1))=0


We calculate terms in parentheses: -(5(c+1)), so:

5(c+1)

We multiply parentheses

5c+5

Back to the equation:

-(5c+5)


We get rid of parentheses

3c^2-6c-5c-5=0

We add all the numbers together, and all the variables

3c^2-11c-5=0

a = 3; b = -11; c = -5;
Δ = b2-4ac
Δ = -112-4·3·(-5)
Δ = 181
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{-(-11)-\sqrt{181}}{2*3}=\frac{11-\sqrt{181}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{181}}{2*3}=\frac{11+\sqrt{181}}{6}
$