-16t^2+42t+36=0

Solution for -16t^2+42t+36=0 equation:

-16t^2+42t+36=0

a = -16; b = 42; c = +36;
Δ = b2-4ac
Δ = 422-4·(-16)·36
Δ = 4068
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:

$\sqrt{\Delta}=\sqrt{4068}=\sqrt{36*113}=\sqrt{36}*\sqrt{113}=6\sqrt{113}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{113}}{2*-16}=\frac{-42-6\sqrt{113}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{113}}{2*-16}=\frac{-42+6\sqrt{113}}{-32}
$