13=8/3t^2

Solution for 13=8/3t^2 equation:

13=8/3t^2

We move all terms to the left:

13-(8/3t^2)=0


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

t!=0/1

t!=0

t∈R


We get rid of parentheses

-8/3t^2+13=0

We multiply all the terms by the denominator


13*3t^2-8=0

Wy multiply elements

39t^2-8=0

a = 39; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·39·(-8)
Δ = 1248
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:

$\sqrt{\Delta}=\sqrt{1248}=\sqrt{16*78}=\sqrt{16}*\sqrt{78}=4\sqrt{78}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{78}}{2*39}=\frac{0-4\sqrt{78}}{78}
=-\frac{4\sqrt{78}}{78}
=-\frac{2\sqrt{78}}{39}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{78}}{2*39}=\frac{0+4\sqrt{78}}{78}
=\frac{4\sqrt{78}}{78}
=\frac{2\sqrt{78}}{39}
$