4(2c+1)=6(c+3)c

Solution for 4(2c+1)=6(c+3)c equation:

4(2c+1)=6(c+3)c

We move all terms to the left:

4(2c+1)-(6(c+3)c)=0

We multiply parentheses

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


We calculate terms in parentheses: -(6(c+3)c), so:

6(c+3)c

We multiply parentheses

6c^2+18c

Back to the equation:

-(6c^2+18c)


We get rid of parentheses

-6c^2+8c-18c+4=0

We add all the numbers together, and all the variables

-6c^2-10c+4=0

a = -6; b = -10; c = +4;
Δ = b2-4ac
Δ = -102-4·(-6)·4
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*-6}=\frac{-4}{-12}
=1/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*-6}=\frac{24}{-12}
=-2
$