13+3/4c=-3/2c

Solution for 13+3/4c=-3/2c equation:

13+3/4c=-3/2c

We move all terms to the left:

13+3/4c-(-3/2c)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R



Domain of the equation: 2c)!=0

c!=0/1

c!=0

c∈R


We get rid of parentheses

3/4c+3/2c+13=0

We calculate fractions


6c/8c^2+12c/8c^2+13=0

We multiply all the terms by the denominator


6c+12c+13*8c^2=0

We add all the numbers together, and all the variables

18c+13*8c^2=0

Wy multiply elements

104c^2+18c=0

a = 104; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·104·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*104}=\frac{-36}{208}
=-9/52
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*104}=\frac{0}{208}
=0
$