0=(-2t^2+1)

Solution for 0=(-2t^2+1) equation:

0=(-2t^2+1)

We move all terms to the left:

0-((-2t^2+1))=0

We add all the numbers together, and all the variables

-((-2t^2+1))=0


We calculate terms in parentheses: -((-2t^2+1)), so:

(-2t^2+1)

We get rid of parentheses

-2t^2+1

Back to the equation:

-(-2t^2+1)


We get rid of parentheses

2t^2-1=0

a = 2; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·2·(-1)
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*2}=\frac{0-2\sqrt{2}}{4}
=-\frac{2\sqrt{2}}{4}
=-\frac{\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*2}=\frac{0+2\sqrt{2}}{4}
=\frac{2\sqrt{2}}{4}
=\frac{\sqrt{2}}{2}
$