21(t)=-16t2+40t+5

Solution for 21(t)=-16t2+40t+5 equation:

21(t)=-16t^2+40t+5

We move all terms to the left:

21(t)-(-16t^2+40t+5)=0

We get rid of parentheses

16t^2-40t+21t-5=0

We add all the numbers together, and all the variables

16t^2-19t-5=0

a = 16; b = -19; c = -5;
Δ = b2-4ac
Δ = -192-4·16·(-5)
Δ = 681
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{-(-19)-\sqrt{681}}{2*16}=\frac{19-\sqrt{681}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{681}}{2*16}=\frac{19+\sqrt{681}}{32}
$