1/5=t^2

Solution for 1/5=t^2 equation:

1/5=t^2

We move all terms to the left:

1/5-(t^2)=0

We add all the numbers together, and all the variables

-1t^2+1/5=0

We multiply all the terms by the denominator


-1t^2*5+1=0

Wy multiply elements

-5t^2+1=0

a = -5; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-5)·1
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*-5}=\frac{0-2\sqrt{5}}{-10}
=-\frac{2\sqrt{5}}{-10}
=-\frac{\sqrt{5}}{-5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*-5}=\frac{0+2\sqrt{5}}{-10}
=\frac{2\sqrt{5}}{-10}
=\frac{\sqrt{5}}{-5}
$