10=40/(t^2)

Solution for 10=40/(t^2) equation:

10=40/(t^2)

We move all terms to the left:

10-(40/(t^2))=0


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

t!=0/1

t!=0

t∈R


We get rid of parentheses

-40/t^2+10=0

We multiply all the terms by the denominator


10*t^2-40=0

We add all the numbers together, and all the variables

10t^2-40=0

a = 10; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·10·(-40)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*10}=\frac{-40}{20}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*10}=\frac{40}{20}
=2
$