3/2t+1=5/4t-5

Solution for 3/2t+1=5/4t-5 equation:

3/2t+1=5/4t-5

We move all terms to the left:

3/2t+1-(5/4t-5)=0


Domain of the equation: 2t!=0

t!=0/2

t!=0

t∈R



Domain of the equation: 4t-5)!=0

t∈R


We get rid of parentheses

3/2t-5/4t+5+1=0

We calculate fractions


12t/8t^2+(-10t)/8t^2+5+1=0

We add all the numbers together, and all the variables

12t/8t^2+(-10t)/8t^2+6=0

We multiply all the terms by the denominator


12t+(-10t)+6*8t^2=0

Wy multiply elements

48t^2+12t+(-10t)=0

We get rid of parentheses

48t^2+12t-10t=0

We add all the numbers together, and all the variables

48t^2+2t=0

a = 48; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·48·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*48}=\frac{-4}{96}
=-1/24
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*48}=\frac{0}{96}
=0
$