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

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

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

We move all terms to the left:

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

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

Wy multiply elements

-2t^2-6t-2=0

a = -2; b = -6; c = -2;
Δ = b2-4ac
Δ = -62-4·(-2)·(-2)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{5}}{2*-2}=\frac{6-2\sqrt{5}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{5}}{2*-2}=\frac{6+2\sqrt{5}}{-4}
$