1.3=5(t)^2

Solution for 1.3=5(t)^2 equation:

1.3=5(t)^2

We move all terms to the left:

1.3-(5(t)^2)=0

determiningTheFunctionDomain
-5t^2+1.3=0

a = -5; b = 0; c = +1.3;
Δ = b2-4ac
Δ = 02-4·(-5)·1.3
Δ = 26
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{26}}{2*-5}=\frac{0-\sqrt{26}}{-10}
=-\frac{\sqrt{}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{26}}{2*-5}=\frac{0+\sqrt{26}}{-10}
=\frac{\sqrt{}}{-10}
$