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

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

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

We move all terms to the left:

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


Domain of the equation: 2t!=0

t!=0/2

t!=0

t∈R



Domain of the equation: 2t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


-1t*2t-3*2t+5-1=0

We add all the numbers together, and all the variables

-1t*2t-3*2t+4=0

Wy multiply elements

-2t^2-6t+4=0

a = -2; b = -6; c = +4;
Δ = b2-4ac
Δ = -62-4·(-2)·4
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{17}}{2*-2}=\frac{6-2\sqrt{17}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{17}}{2*-2}=\frac{6+2\sqrt{17}}{-4}
$