1(t)=-16t^2+8t+24

Solution for 1(t)=-16t^2+8t+24 equation:

1(t)=-16t^2+8t+24

We move all terms to the left:

1(t)-(-16t^2+8t+24)=0

We add all the numbers together, and all the variables

-(-16t^2+8t+24)+t=0

We get rid of parentheses

16t^2-8t+t-24=0

We add all the numbers together, and all the variables

16t^2-7t-24=0

a = 16; b = -7; c = -24;
Δ = b2-4ac
Δ = -72-4·16·(-24)
Δ = 1585
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{-(-7)-\sqrt{1585}}{2*16}=\frac{7-\sqrt{1585}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{1585}}{2*16}=\frac{7+\sqrt{1585}}{32}
$