25=5t^2+25t

Solution for 25=5t^2+25t equation:

25=5t^2+25t

We move all terms to the left:

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

We get rid of parentheses

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

a = -5; b = -25; c = +25;
Δ = b2-4ac
Δ = -252-4·(-5)·25
Δ = 1125
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:

$\sqrt{\Delta}=\sqrt{1125}=\sqrt{225*5}=\sqrt{225}*\sqrt{5}=15\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-15\sqrt{5}}{2*-5}=\frac{25-15\sqrt{5}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+15\sqrt{5}}{2*-5}=\frac{25+15\sqrt{5}}{-10}
$