13=-16t^2+21t+7

Solution for 13=-16t^2+21t+7 equation:

13=-16t^2+21t+7

We move all terms to the left:

13-(-16t^2+21t+7)=0

We get rid of parentheses

16t^2-21t-7+13=0

We add all the numbers together, and all the variables

16t^2-21t+6=0

a = 16; b = -21; c = +6;
Δ = b2-4ac
Δ = -212-4·16·6
Δ = 57
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{-(-21)-\sqrt{57}}{2*16}=\frac{21-\sqrt{57}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{57}}{2*16}=\frac{21+\sqrt{57}}{32}
$