0(t)=-16t^2+88t+48

Solution for 0(t)=-16t^2+88t+48 equation:

0(t)=-16t^2+88t+48

We move all terms to the left:

0(t)-(-16t^2+88t+48)=0

We add all the numbers together, and all the variables

-(-16t^2+88t+48)+t=0

We get rid of parentheses

16t^2-88t+t-48=0

We add all the numbers together, and all the variables

16t^2-87t-48=0

a = 16; b = -87; c = -48;
Δ = b2-4ac
Δ = -872-4·16·(-48)
Δ = 10641
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{-(-87)-\sqrt{10641}}{2*16}=\frac{87-\sqrt{10641}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-87)+\sqrt{10641}}{2*16}=\frac{87+\sqrt{10641}}{32}
$