1/5t2-9=0

Solution for 1/5t2-9=0 equation:

1/5t^2-9=0


Domain of the equation: 5t^2!=0

t^2!=0/5

t^2!=√0

t!=0

t∈R


We multiply all the terms by the denominator


-9*5t^2+1=0

Wy multiply elements

-45t^2+1=0

a = -45; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-45)·1
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*-45}=\frac{0-6\sqrt{5}}{-90}
=-\frac{6\sqrt{5}}{-90}
=-\frac{\sqrt{5}}{-15}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*-45}=\frac{0+6\sqrt{5}}{-90}
=\frac{6\sqrt{5}}{-90}
=\frac{\sqrt{5}}{-15}
$