(6-t^2)/(6+t^2)^2=0

Solution for (6-t^2)/(6+t^2)^2=0 equation:

(6-t^2)/(6+t^2)^2=0


Domain of the equation: (6+t^2)^2!=0

t∈R


We multiply all the terms by the denominator


(6-t^2)=0

We get rid of parentheses

-t^2+6=0

We add all the numbers together, and all the variables

-1t^2+6=0

a = -1; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-1)·6
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-1}=\frac{0-2\sqrt{6}}{-2}
=-\frac{2\sqrt{6}}{-2}
=-\frac{\sqrt{6}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-1}=\frac{0+2\sqrt{6}}{-2}
=\frac{2\sqrt{6}}{-2}
=\frac{\sqrt{6}}{-1}
$