16(t)=-16(t)^2+20

Solution for 16(t)=-16(t)^2+20 equation:

16(t)=-16(t)^2+20

We move all terms to the left:

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

determiningTheFunctionDomain
-(-16t^2+20)+16t=0

We get rid of parentheses

16t^2+16t-20=0

a = 16; b = 16; c = -20;
Δ = b2-4ac
Δ = 162-4·16·(-20)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{6}}{2*16}=\frac{-16-16\sqrt{6}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{6}}{2*16}=\frac{-16+16\sqrt{6}}{32}
$