21/2+3/5t=1/4t

Solution for 21/2+3/5t=1/4t equation:

21/2+3/5t=1/4t

We move all terms to the left:

21/2+3/5t-(1/4t)=0


Domain of the equation: 5t!=0

t!=0/5

t!=0

t∈R



Domain of the equation: 4t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

3/5t-(+1/4t)+21/2=0

We get rid of parentheses

3/5t-1/4t+21/2=0

We calculate fractions


1680t^2/80t^2+48t/80t^2+(-20t)/80t^2=0

We multiply all the terms by the denominator


1680t^2+48t+(-20t)=0

We get rid of parentheses

1680t^2+48t-20t=0

We add all the numbers together, and all the variables

1680t^2+28t=0

a = 1680; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·1680·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*1680}=\frac{-56}{3360}
=-1/60
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*1680}=\frac{0}{3360}
=0
$