13=3+25t-5t^2

Solution for 13=3+25t-5t^2 equation:

13=3+25t-5t^2

We move all terms to the left:

13-(3+25t-5t^2)=0

We get rid of parentheses

5t^2-25t-3+13=0

We add all the numbers together, and all the variables

5t^2-25t+10=0

a = 5; b = -25; c = +10;
Δ = b2-4ac
Δ = -252-4·5·10
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{17}}{2*5}=\frac{25-5\sqrt{17}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{17}}{2*5}=\frac{25+5\sqrt{17}}{10}
$