0=23.16(t)+4.9(t^2)

Solution for 0=23.16(t)+4.9(t^2) equation:

0=23.16(t)+4.9(t^2)

We move all terms to the left:

0-(23.16(t)+4.9(t^2))=0

We add all the numbers together, and all the variables

-(23.16t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-23.16t=0

a = -4.9; b = -23.16; c = 0;
Δ = b2-4ac
Δ = -23.162-4·(-4.9)·0
Δ = 536.3856
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{-(-23.16)-\sqrt{536.3856}}{2*-4.9}=\frac{23.16-\sqrt{536.3856}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23.16)+\sqrt{536.3856}}{2*-4.9}=\frac{23.16+\sqrt{536.3856}}{-9.8}
$