-16t^2+54t+40=50

Solution for -16t^2+54t+40=50 equation:

-16t^2+54t+40=50

We move all terms to the left:

-16t^2+54t+40-(50)=0

We add all the numbers together, and all the variables

-16t^2+54t-10=0

a = -16; b = 54; c = -10;
Δ = b2-4ac
Δ = 542-4·(-16)·(-10)
Δ = 2276
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:

$\sqrt{\Delta}=\sqrt{2276}=\sqrt{4*569}=\sqrt{4}*\sqrt{569}=2\sqrt{569}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-2\sqrt{569}}{2*-16}=\frac{-54-2\sqrt{569}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+2\sqrt{569}}{2*-16}=\frac{-54+2\sqrt{569}}{-32}
$