25=2t^2

Solution for 25=2t^2 equation:

25=2t^2

We move all terms to the left:

25-(2t^2)=0

a = -2; b = 0; c = +25;
Δ = b2-4ac
Δ = 02-4·(-2)·25
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*-2}=\frac{0-10\sqrt{2}}{-4}
=-\frac{10\sqrt{2}}{-4}
=-\frac{5\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*-2}=\frac{0+10\sqrt{2}}{-4}
=\frac{10\sqrt{2}}{-4}
=\frac{5\sqrt{2}}{-2}
$