3+25t-5t^2=13

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

3+25t-5t^2=13

We move all terms to the left:

3+25t-5t^2-(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}
$