t^2=129

Solution for t^2=129 equation:

t^2=129

We move all terms to the left:

t^2-(129)=0

a = 1; b = 0; c = -129;
Δ = b2-4ac
Δ = 02-4·1·(-129)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{129}}{2*1}=\frac{0-2\sqrt{129}}{2}
=-\frac{2\sqrt{129}}{2}
=-\sqrt{129}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{129}}{2*1}=\frac{0+2\sqrt{129}}{2}
=\frac{2\sqrt{129}}{2}
=\sqrt{129}
$