49=490-16t^2

Solution for 49=490-16t^2 equation:

49=490-16t^2

We move all terms to the left:

49-(490-16t^2)=0

We get rid of parentheses

16t^2-490+49=0

We add all the numbers together, and all the variables

16t^2-441=0

a = 16; b = 0; c = -441;
Δ = b2-4ac
Δ = 02-4·16·(-441)
Δ = 28224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{28224}=168$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-168}{2*16}=\frac{-168}{32}
=-5+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+168}{2*16}=\frac{168}{32}
=5+1/4
$