1.1=4.9t^2

Solution for 1.1=4.9t^2 equation:

1.1=4.9t^2

We move all terms to the left:

1.1-(4.9t^2)=0

We get rid of parentheses

-4.9t^2+1.1=0

a = -4.9; b = 0; c = +1.1;
Δ = b2-4ac
Δ = 02-4·(-4.9)·1.1
Δ = 21.56
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{-(0)-\sqrt{21.56}}{2*-4.9}=\frac{0-\sqrt{21.56}}{-9.8}
=-\frac{\sqrt{}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{21.56}}{2*-4.9}=\frac{0+\sqrt{21.56}}{-9.8}
=\frac{\sqrt{}}{-9.8}
$