-16t^2+120t+10=t

Solution for -16t^2+120t+10=t equation:

-16t^2+120t+10=t

We move all terms to the left:

-16t^2+120t+10-(t)=0

We add all the numbers together, and all the variables

-16t^2+119t+10=0

a = -16; b = 119; c = +10;
Δ = b2-4ac
Δ = 1192-4·(-16)·10
Δ = 14801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14801}=\sqrt{361*41}=\sqrt{361}*\sqrt{41}=19\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(119)-19\sqrt{41}}{2*-16}=\frac{-119-19\sqrt{41}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(119)+19\sqrt{41}}{2*-16}=\frac{-119+19\sqrt{41}}{-32}
$