13=-16t^2+25t+20

Solution for 13=-16t^2+25t+20 equation:

13=-16t^2+25t+20

We move all terms to the left:

13-(-16t^2+25t+20)=0

We get rid of parentheses

16t^2-25t-20+13=0

We add all the numbers together, and all the variables

16t^2-25t-7=0

a = 16; b = -25; c = -7;
Δ = b2-4ac
Δ = -252-4·16·(-7)
Δ = 1073
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{1073}}{2*16}=\frac{25-\sqrt{1073}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{1073}}{2*16}=\frac{25+\sqrt{1073}}{32}
$