22t=16t^2

Solution for 22t=16t^2 equation:

22t=16t^2

We move all terms to the left:

22t-(16t^2)=0

determiningTheFunctionDomain
-16t^2+22t=0

a = -16; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-16)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-16}=\frac{-44}{-32}
=1+3/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-16}=\frac{0}{-32}
=0
$