0.12=5t^2

Solution for 0.12=5t^2 equation:

0.12=5t^2

We move all terms to the left:

0.12-(5t^2)=0

a = -5; b = 0; c = +0.12;
Δ = b2-4ac
Δ = 02-4·(-5)·0.12
Δ = 2.4
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{2.4}}{2*-5}=\frac{0-\sqrt{2.4}}{-10}
=-\frac{\sqrt{}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{2.4}}{2*-5}=\frac{0+\sqrt{2.4}}{-10}
=\frac{\sqrt{}}{-10}
$