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

Solution for t=-16t^2+-2t+24 equation:

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

We move all terms to the left:

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

We use the square of the difference formula

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2+3t-24=0

a = 16; b = 3; c = -24;
Δ = b2-4ac
Δ = 32-4·16·(-24)
Δ = 1545
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{-(3)-\sqrt{1545}}{2*16}=\frac{-3-\sqrt{1545}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1545}}{2*16}=\frac{-3+\sqrt{1545}}{32}
$