-t^2+2=t0

Solution for -t^2+2=t0 equation:

-t^2+2=t0

We move all terms to the left:

-t^2+2-(t0)=0

We add all the numbers together, and all the variables

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

a = -1; b = -1; c = +2;
Δ = b2-4ac
Δ = -12-4·(-1)·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*-1}=\frac{-2}{-2}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*-1}=\frac{4}{-2}
=-2
$