2(2c+4)=-(3c+6)/4

Solution for 2(2c+4)=-(3c+6)/4 equation:

2(2c+4)=-(3c+6)/4

We move all terms to the left:

2(2c+4)-(-(3c+6)/4)=0

We multiply parentheses

4c-(-(3c+6)/4)+8=0

We multiply all the terms by the denominator


4c*4)-(-(3c+6)+8*4)=0

We add all the numbers together, and all the variables

4c*4)-(-(3c+6)=0

Wy multiply elements

16c^2-(3c+6)=0

We get rid of parentheses

16c^2-3c-6=0

a = 16; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·16·(-6)
Δ = 393
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{-(-3)-\sqrt{393}}{2*16}=\frac{3-\sqrt{393}}{32}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{393}}{2*16}=\frac{3+\sqrt{393}}{32}
$