25=10t^2-5t

Solution for 25=10t^2-5t equation:

25=10t^2-5t

We move all terms to the left:

25-(10t^2-5t)=0

We get rid of parentheses

-10t^2+5t+25=0

a = -10; b = 5; c = +25;
Δ = b2-4ac
Δ = 52-4·(-10)·25
Δ = 1025
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:

$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{41}}{2*-10}=\frac{-5-5\sqrt{41}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{41}}{2*-10}=\frac{-5+5\sqrt{41}}{-20}
$