t^2=0,04

Solution for t^2=0,04 equation:

t^2=0.04

We move all terms to the left:

t^2-(0.04)=0

We add all the numbers together, and all the variables

t^2-0.04=0

a = 1; b = 0; c = -0.04;
Δ = b2-4ac
Δ = 02-4·1·(-0.04)
Δ = 0.16
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{0.16}}{2*1}=\frac{0-\sqrt{0.16}}{2}
=-\frac{\sqrt{}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.16}}{2*1}=\frac{0+\sqrt{0.16}}{2}
=\frac{\sqrt{}}{2}
$