-16t^2+63t+42=0

Solution for -16t^2+63t+42=0 equation:

-16t^2+63t+42=0

a = -16; b = 63; c = +42;
Δ = b2-4ac
Δ = 632-4·(-16)·42
Δ = 6657
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{-(63)-\sqrt{6657}}{2*-16}=\frac{-63-\sqrt{6657}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+\sqrt{6657}}{2*-16}=\frac{-63+\sqrt{6657}}{-32}
$