T=-16(2t-3)(t+1)

Solution for T=-16(2t-3)(t+1) equation:

=-16(2T-3)(T+1)

We move all terms to the left:

-(-16(2T-3)(T+1))=0

We multiply parentheses ..

-(-16(+2T^2+2T-3T-3))=0


We calculate terms in parentheses: -(-16(+2T^2+2T-3T-3)), so:

-16(+2T^2+2T-3T-3)

We multiply parentheses

-32T^2-32T+48T+48

We add all the numbers together, and all the variables

-32T^2+16T+48

Back to the equation:

-(-32T^2+16T+48)


We get rid of parentheses

32T^2-16T-48=0

a = 32; b = -16; c = -48;
Δ = b2-4ac
Δ = -162-4·32·(-48)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-80}{2*32}=\frac{-64}{64}
=-1
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+80}{2*32}=\frac{96}{64}
=1+1/2
$