4/t-3=t+1/t-3

Solution for 4/t-3=t+1/t-3 equation:

4/t-3=t+1/t-3

We move all terms to the left:

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


Domain of the equation: t!=0

t∈R



Domain of the equation: t-3)!=0

t∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


-t*t+3*t-3*t+4-1=0

We add all the numbers together, and all the variables

-t*t+3=0

Wy multiply elements

-1t^2+3=0

a = -1; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-1)·3
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-1}=\frac{0-2\sqrt{3}}{-2}
=-\frac{2\sqrt{3}}{-2}
=-\frac{\sqrt{3}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-1}=\frac{0+2\sqrt{3}}{-2}
=\frac{2\sqrt{3}}{-2}
=\frac{\sqrt{3}}{-1}
$