42=16t^2

Solution for 42=16t^2 equation:

42=16t^2

We move all terms to the left:

42-(16t^2)=0

a = -16; b = 0; c = +42;
Δ = b2-4ac
Δ = 02-4·(-16)·42
Δ = 2688
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:

$\sqrt{\Delta}=\sqrt{2688}=\sqrt{64*42}=\sqrt{64}*\sqrt{42}=8\sqrt{42}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{42}}{2*-16}=\frac{0-8\sqrt{42}}{-32}
=-\frac{8\sqrt{42}}{-32}
=-\frac{\sqrt{42}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{42}}{2*-16}=\frac{0+8\sqrt{42}}{-32}
=\frac{8\sqrt{42}}{-32}
=\frac{\sqrt{42}}{-4}
$