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

Solution for (2/t-2)-(t+1/t+4)=0 equation:

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


Domain of the equation: t-2)!=0

t∈R



Domain of the equation: t+4)!=0

t∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

-6t-t*t+1=0

Wy multiply elements

-1t^2-6t+1=0

a = -1; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·(-1)·1
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{10}}{2*-1}=\frac{6-2\sqrt{10}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{10}}{2*-1}=\frac{6+2\sqrt{10}}{-2}
$