3/4c+8=c+12

Solution for 3/4c+8=c+12 equation:

3/4c+8=c+12

We move all terms to the left:

3/4c+8-(c+12)=0


Domain of the equation: 4c!=0

c!=0/4

c!=0

c∈R


We get rid of parentheses

3/4c-c-12+8=0

We multiply all the terms by the denominator


-c*4c-12*4c+8*4c+3=0

Wy multiply elements

-4c^2-48c+32c+3=0

We add all the numbers together, and all the variables

-4c^2-16c+3=0

a = -4; b = -16; c = +3;
Δ = b2-4ac
Δ = -162-4·(-4)·3
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{19}}{2*-4}=\frac{16-4\sqrt{19}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{19}}{2*-4}=\frac{16+4\sqrt{19}}{-8}
$