1/4t-1/3t=12

Solution for 1/4t-1/3t=12 equation:

1/4t-1/3t=12

We move all terms to the left:

1/4t-1/3t-(12)=0


Domain of the equation: 4t!=0

t!=0/4

t!=0

t∈R



Domain of the equation: 3t!=0

t!=0/3

t!=0

t∈R


We calculate fractions


3t/12t^2+(-4t)/12t^2-12=0

We multiply all the terms by the denominator


3t+(-4t)-12*12t^2=0

Wy multiply elements

-144t^2+3t+(-4t)=0

We get rid of parentheses

-144t^2+3t-4t=0

We add all the numbers together, and all the variables

-144t^2-1t=0

a = -144; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-144)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-144}=\frac{0}{-288}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-144}=\frac{2}{-288}
=-1/144
$