5c+3c(c+4)=36

Solution for 5c+3c(c+4)=36 equation:

5c+3c(c+4)=36

We move all terms to the left:

5c+3c(c+4)-(36)=0

We multiply parentheses

3c^2+5c+12c-36=0

We add all the numbers together, and all the variables

3c^2+17c-36=0

a = 3; b = 17; c = -36;
Δ = b2-4ac
Δ = 172-4·3·(-36)
Δ = 721
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{-(17)-\sqrt{721}}{2*3}=\frac{-17-\sqrt{721}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{721}}{2*3}=\frac{-17+\sqrt{721}}{6}
$