4c(c-3)+3c=16

Solution for 4c(c-3)+3c=16 equation:

4c(c-3)+3c=16

We move all terms to the left:

4c(c-3)+3c-(16)=0

We add all the numbers together, and all the variables

3c+4c(c-3)-16=0

We multiply parentheses

4c^2+3c-12c-16=0

We add all the numbers together, and all the variables

4c^2-9c-16=0

a = 4; b = -9; c = -16;
Δ = b2-4ac
Δ = -92-4·4·(-16)
Δ = 337
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{337}}{2*4}=\frac{9-\sqrt{337}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{337}}{2*4}=\frac{9+\sqrt{337}}{8}
$