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

Solution for (4(t^2-6))/((t^2+6)^2)=0 equation:

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


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

t∈R


We multiply all the terms by the denominator


(4(t^2-6))=0


We calculate terms in parentheses: +(4(t^2-6)), so:

4(t^2-6)

We multiply parentheses

4t^2-24

Back to the equation:

+(4t^2-24)


We get rid of parentheses

4t^2-24=0

a = 4; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·4·(-24)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*4}=\frac{0-8\sqrt{6}}{8}
=-\frac{8\sqrt{6}}{8}
=-\sqrt{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*4}=\frac{0+8\sqrt{6}}{8}
=\frac{8\sqrt{6}}{8}
=\sqrt{6}
$