5/2(t)+t=3+3/2(t)

Solution for 5/2(t)+t=3+3/2(t) equation:

5/2(t)+t=3+3/2(t)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

t*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}
$