t=8/t=2

Solution for t=8/t=2 equation:

t=8/t=2

We move all terms to the left:

t-(8/t)=0


Domain of the equation: t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

t-(+8/t)=0

We get rid of parentheses

t-8/t=0

We multiply all the terms by the denominator


t*t-8=0

Wy multiply elements

t^2-8=0

a = 1; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·1·(-8)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*1}=\frac{0-4\sqrt{2}}{2}
=-\frac{4\sqrt{2}}{2}
=-2\sqrt{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*1}=\frac{0+4\sqrt{2}}{2}
=\frac{4\sqrt{2}}{2}
=2\sqrt{2}
$