1/2+2/5t-4=1/5t+t

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

1/2+2/5t-4=1/5t+t

We move all terms to the left:

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


Domain of the equation: 5t!=0

t!=0/5

t!=0

t∈R



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

t∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


-t+()/10t+5t/10t-4=0

We add all the numbers together, and all the variables

-1t+()/10t+5t/10t-4=0

We multiply all the terms by the denominator


-1t*10t+5t-4*10t+()=0

We add all the numbers together, and all the variables

5t-1t*10t-4*10t=0

Wy multiply elements

-10t^2+5t-40t=0

We add all the numbers together, and all the variables

-10t^2-35t=0

a = -10; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·(-10)·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*-10}=\frac{0}{-20}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*-10}=\frac{70}{-20}
=-3+1/2
$