122(t)=-16t2+64t+48,

Solution for 122(t)=-16t2+64t+48, equation:

122(t)=-16t^2+64t+48.

We move all terms to the left:

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

We get rid of parentheses

16t^2-64t+122t-48.=0

We add all the numbers together, and all the variables

16t^2+58t-48=0

a = 16; b = 58; c = -48;
Δ = b2-4ac
Δ = 582-4·16·(-48)
Δ = 6436
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:

$\sqrt{\Delta}=\sqrt{6436}=\sqrt{4*1609}=\sqrt{4}*\sqrt{1609}=2\sqrt{1609}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-2\sqrt{1609}}{2*16}=\frac{-58-2\sqrt{1609}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+2\sqrt{1609}}{2*16}=\frac{-58+2\sqrt{1609}}{32}
$