1/3t^2+19=83

Solution for 1/3t^2+19=83 equation:

1/3t^2+19=83

We move all terms to the left:

1/3t^2+19-(83)=0


Domain of the equation: 3t^2!=0

t^2!=0/3

t^2!=√0

t!=0

t∈R


We add all the numbers together, and all the variables

1/3t^2-64=0

We multiply all the terms by the denominator


-64*3t^2+1=0

Wy multiply elements

-192t^2+1=0

a = -192; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-192)·1
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*-192}=\frac{0-16\sqrt{3}}{-384}
=-\frac{16\sqrt{3}}{-384}
=-\frac{\sqrt{3}}{-24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*-192}=\frac{0+16\sqrt{3}}{-384}
=\frac{16\sqrt{3}}{-384}
=\frac{\sqrt{3}}{-24}
$