4c(c+12)=2c+18

Solution for 4c(c+12)=2c+18 equation:

4c(c+12)=2c+18

We move all terms to the left:

4c(c+12)-(2c+18)=0

We multiply parentheses

4c^2+48c-(2c+18)=0

We get rid of parentheses

4c^2+48c-2c-18=0

We add all the numbers together, and all the variables

4c^2+46c-18=0

a = 4; b = 46; c = -18;
Δ = b2-4ac
Δ = 462-4·4·(-18)
Δ = 2404
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:

$\sqrt{\Delta}=\sqrt{2404}=\sqrt{4*601}=\sqrt{4}*\sqrt{601}=2\sqrt{601}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-2\sqrt{601}}{2*4}=\frac{-46-2\sqrt{601}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+2\sqrt{601}}{2*4}=\frac{-46+2\sqrt{601}}{8}
$