62=-16t^2-22t+100

Solution for 62=-16t^2-22t+100 equation:

62=-16t^2-22t+100

We move all terms to the left:

62-(-16t^2-22t+100)=0

We get rid of parentheses

16t^2+22t-100+62=0

We add all the numbers together, and all the variables

16t^2+22t-38=0

a = 16; b = 22; c = -38;
Δ = b2-4ac
Δ = 222-4·16·(-38)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-54}{2*16}=\frac{-76}{32}
=-2+3/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+54}{2*16}=\frac{32}{32}
=1
$