64(t)=-16t^2+48+0

Solution for 64(t)=-16t^2+48+0 equation:

64(t)=-16t^2+48+0

We move all terms to the left:

64(t)-(-16t^2+48+0)=0

We get rid of parentheses

16t^2+64t-48-0=0

We add all the numbers together, and all the variables

16t^2+64t-48=0

a = 16; b = 64; c = -48;
Δ = b2-4ac
Δ = 642-4·16·(-48)
Δ = 7168
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:

$\sqrt{\Delta}=\sqrt{7168}=\sqrt{1024*7}=\sqrt{1024}*\sqrt{7}=32\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-32\sqrt{7}}{2*16}=\frac{-64-32\sqrt{7}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+32\sqrt{7}}{2*16}=\frac{-64+32\sqrt{7}}{32}
$