0=-16t2+42t+5

Solution for 0=-16t2+42t+5 equation:

0=-16t^2+42t+5

We move all terms to the left:

0-(-16t^2+42t+5)=0

We add all the numbers together, and all the variables

-(-16t^2+42t+5)=0

We get rid of parentheses

16t^2-42t-5=0

a = 16; b = -42; c = -5;
Δ = b2-4ac
Δ = -422-4·16·(-5)
Δ = 2084
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:

$\sqrt{\Delta}=\sqrt{2084}=\sqrt{4*521}=\sqrt{4}*\sqrt{521}=2\sqrt{521}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{521}}{2*16}=\frac{42-2\sqrt{521}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{521}}{2*16}=\frac{42+2\sqrt{521}}{32}
$