-16t^2+4t+22=t

Solution for -16t^2+4t+22=t equation:

-16t^2+4t+22=t

We move all terms to the left:

-16t^2+4t+22-(t)=0

We add all the numbers together, and all the variables

-16t^2+3t+22=0

a = -16; b = 3; c = +22;
Δ = b2-4ac
Δ = 32-4·(-16)·22
Δ = 1417
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{1417}}{2*-16}=\frac{-3-\sqrt{1417}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1417}}{2*-16}=\frac{-3+\sqrt{1417}}{-32}
$