4c(c+3)=-8

Solution for 4c(c+3)=-8 equation:

4c(c+3)=-8

We move all terms to the left:

4c(c+3)-(-8)=0

We add all the numbers together, and all the variables

4c(c+3)+8=0

We multiply parentheses

4c^2+12c+8=0

a = 4; b = 12; c = +8;
Δ = b2-4ac
Δ = 122-4·4·8
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*4}=\frac{-16}{8}
=-2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*4}=\frac{-8}{8}
=-1
$