0=-4.9t^2+23

Solution for 0=-4.9t^2+23 equation:

0=-4.9t^2+23

We move all terms to the left:

0-(-4.9t^2+23)=0

We add all the numbers together, and all the variables

-(-4.9t^2+23)=0

We get rid of parentheses

4.9t^2-23=0

a = 4.9; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·4.9·(-23)
Δ = 450.8
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{450.8}}{2*4.9}=\frac{0-\sqrt{450.8}}{9.8}
=-\frac{\sqrt{}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{450.8}}{2*4.9}=\frac{0+\sqrt{450.8}}{9.8}
=\frac{\sqrt{}}{9.8}
$