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

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

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

We move all terms to the left:

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


Domain of the equation: 2)t!=0

t!=0/1

t!=0

t∈R



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

We move all terms containing t to the left, all other terms to the right

-t)!=-5

t!=-5/1

t!=-5

t∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We multiply all the terms by the denominator


t^2*5+1t*5-3+4=0

We add all the numbers together, and all the variables

t^2*5+1t*5+1=0

Wy multiply elements

5t^2+5t+1=0

a = 5; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·5·1
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{5}}{2*5}=\frac{-5-\sqrt{5}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{5}}{2*5}=\frac{-5+\sqrt{5}}{10}
$