-t^2+245=0

Solution for -t^2+245=0 equation:

-t^2+245=0

We add all the numbers together, and all the variables

-1t^2+245=0

a = -1; b = 0; c = +245;
Δ = b2-4ac
Δ = 02-4·(-1)·245
Δ = 980
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:

$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{5}}{2*-1}=\frac{0-14\sqrt{5}}{-2}
=-\frac{14\sqrt{5}}{-2}
=-\frac{7\sqrt{5}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{5}}{2*-1}=\frac{0+14\sqrt{5}}{-2}
=\frac{14\sqrt{5}}{-2}
=\frac{7\sqrt{5}}{-1}
$