-16t^2+53t+40=50

Solution for -16t^2+53t+40=50 equation:

-16t^2+53t+40=50

We move all terms to the left:

-16t^2+53t+40-(50)=0

We add all the numbers together, and all the variables

-16t^2+53t-10=0

a = -16; b = 53; c = -10;
Δ = b2-4ac
Δ = 532-4·(-16)·(-10)
Δ = 2169
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:

$\sqrt{\Delta}=\sqrt{2169}=\sqrt{9*241}=\sqrt{9}*\sqrt{241}=3\sqrt{241}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-3\sqrt{241}}{2*-16}=\frac{-53-3\sqrt{241}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+3\sqrt{241}}{2*-16}=\frac{-53+3\sqrt{241}}{-32}
$