-2/3c+3=2c-5

Solution for -2/3c+3=2c-5 equation:

-2/3c+3=2c-5

We move all terms to the left:

-2/3c+3-(2c-5)=0


Domain of the equation: 3c!=0

c!=0/3

c!=0

c∈R


We get rid of parentheses

-2/3c-2c+5+3=0

We multiply all the terms by the denominator


-2c*3c+5*3c+3*3c-2=0

Wy multiply elements

-6c^2+15c+9c-2=0

We add all the numbers together, and all the variables

-6c^2+24c-2=0

a = -6; b = 24; c = -2;
Δ = b2-4ac
Δ = 242-4·(-6)·(-2)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{33}}{2*-6}=\frac{-24-4\sqrt{33}}{-12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{33}}{2*-6}=\frac{-24+4\sqrt{33}}{-12}
$