-4.9t^2+11=0

Solution for -4.9t^2+11=0 equation:

-4.9t^2+11=0

a = -4.9; b = 0; c = +11;
Δ = b2-4ac
Δ = 02-4·(-4.9)·11
Δ = 215.6
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{215.6}}{2*-4.9}=\frac{0-\sqrt{215.6}}{-9.8}
=-\frac{\sqrt{}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{215.6}}{2*-4.9}=\frac{0+\sqrt{215.6}}{-9.8}
=\frac{\sqrt{}}{-9.8}
$