t^2=68

Solution for t^2=68 equation:

t^2=68

We move all terms to the left:

t^2-(68)=0

a = 1; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·1·(-68)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{17}}{2*1}=\frac{0-4\sqrt{17}}{2}
=-\frac{4\sqrt{17}}{2}
=-2\sqrt{17}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{17}}{2*1}=\frac{0+4\sqrt{17}}{2}
=\frac{4\sqrt{17}}{2}
=2\sqrt{17}
$